Mathematica solve equation in terms of variable. I've tried the following.

Mathematica solve equation in terms of variable why mathematica Holds the Solve on the set of equations in the diagonal a matrix? Note: Since this question has a much simpler structure than the question linked by Artes, these tricks work for the current case (with a single-variable expression to be solved), but not for the more general case in the linked Q/A. Solving a system of linear equations. Numerically solving an equation in which variable is the upper limit of an integral [duplicate] Ask Question Asked 8 years, also dont litter mathematica with superflous real literals. My code is below. In general, the two equations have no common roots, so Solve returns an empty set, { }. This means that given a logical statement (equations, inequalities or I have a system of 4 linear equations in terms of variables that I have obtained from solving previous systems, How to solve an equation system in Mathematica 9. But the solution Mathematica gives does not have y' or dy/dx in it (like when I solved it by hand). Fuhrman - You have four unknowns so Solve needs four equations to determine them. Simply use the variable you want to substitute as the first When equating any of the above equations to zero, each term should vanish separately, so that eq1=0 gives a[3] - a[4] But I want Mathematica to do the solving step for me and I don't want to specify for what variables it should solve for, It doesn't matter if It solves for a[1], a[3], a[4] as a function of a[2] I have an equation which I want to rearrange so one variable is on the left side by itself. Solve [ eqns , vars , elims ] find solutions for Solve is the Mathematica function used for symbolically solving a polynomial equation or set of equations. I've been struggling to solve equations in two variables containing constant symbolic $ are treated as variables by Mathematica, and a solution is provided also for them So how can I get the solution for x and y (without them been coupled) in terms of the constant parameters? equation-solving; parametric-functions; Share. In certain cases, a different ordering can yield different solutions that satisfy the equation or system of equations to be solved. Learn more about: Equation solving; Tips for entering queries. Thanks for contributing an answer to Mathematica Stack Exchange! $\begingroup$ Usually in Quantum Field Theory, when we write the equations for the $\beta$-function, we know or can get the answer in terms of local d. Another way of doing this is to use pure functions and map to perform I'm fairly new to Mathematica, and I'm trying to solve a complicated equation with multiple variables. We can hijack this by just saying your expression is equal to a new variable—now it's an equation: eq1 = (expr == (81 py^4)/(16 (wa + 2 wb)^4)) (Note the difference between = and ==. Using your unedited set of equations, so that we know what variables to solve for: FullSimplify[ r /. I use the code I am trying to solve a linear system of four equations with Mathematica. A and B are square, the elements of A and B don't commute (i. Use NSolve or FindRoot to numerically solve algebraic equations. x == b. x y+y^2-1+z y x-24*z+Tan[x] by this I imply that Mathematica will need to automatically treat z+Cos[z] (instead of only z) As pointed out in the comments, you can't solve an expression. I can understand as some of the equations are quite messy-- in my opinion at least (I'm not a mathematician). I'm trying to solve an equation of this type on integers (10 - x) (6 - y) (4 - z) (3 - w) == x*y*z*w for 1 < = x < In fact I want to solve an equation with more than 20 variables. 65 to 7. Here's my one line code in wolfram Mathematica. I am a novice to Mathematica; is anyone able to do NSolve for x on Mathematica and get a numerical result? I want to get the formula for E1 in terms of the other variables and constants. Commented Jan 5, Getting Mathematica to $\begingroup$ Welcome to Mathematica. If you want to solve an equation with some restriction on the variable domain (field in which the unknown lives), then you can supplement the constraint to the Solve function using the && operator. In Mathematica, calling Solve, returns a list of rules, e. I have tried to get those using This tells you that for a=0 and b=0 the equation a x + b == 0 is always satisfied, regardless of the value of x. 2) Take the tour!3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. This will turn the list of rules returned by Solve into a list of assignments to the variables. In your case it becomes : Solve[x + y == 2 x, y] that gives: {{y -> x}} Using Solve certainly works here, and it gives the correct answer (which is that Abs[x] == Sqrt[5 - y^3]). Original equation is on picture below. Provide details and share your research! But avoid Asking for help, clarification, or responding to other I have a system of ten complex equations, with four complex variables u1,u2,v1,v2 and two real variables θ1, θ2. Add a comment | 2 I am trying to get an equation in terms of variable z by eliminating variables g0 and g1 but. Solve[((30 e i L^2 + k L^5) p0)/(48 e i (3 e i + k L^3))* x^2 - ((L (12 e i + k L^3) p0)/ Thanks for contributing an answer to Mathematica Stack Exchange! $\begingroup$ @azerbajdzan, you cannot claim that there is "a bug" and that "it is really embarrassing for Mathematica" if you are expecting functions to work differently than explained in the documentation. Here the solution expresses one variable in terms of another: To use one of these solutions (here the first one is shown), use [[]] (the short form of Part) to extract it from the list of solutions and use /. To begin, cast the two equations into the proper syntax by replacing = by ==. Since the code is in Mathematica format and it is too long to describe, I attached this mathematica code file to this question post. $\endgroup$ – Of course, we can limit the domain to Integers. I need a way to solve an equation like this using Mathematica to save time. I just want it to be simply rearranged to be in terms of one variable on one side. Simply separate each substitution with a comma. 97}^x \frac{0. 7 & 8 I'm going to suggest how to deal with it in earlier versions. Maybe it helps nevertheless. Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. Its syntax is Solve[eqns, vars], where eqns is your equation or set of equations and vars are the variable(s) in the equation(s). Sometimes it keeps running until my computer crashes or it I think this should be easy to numerically solve in Mathematica, but for some reason I'm not finding the correct way to do it. I am trying to solve for variable kp and would like to know how the value of kp differs for different values of x, y and z which is defined. What statement should I write such that if applied to a x^2 + b x + c y returns a. $\endgroup$ I'm trying to solve a problem in Mathematica where I have 4 unknown variables and 4 equations, but it doesn't quite work for me. 65 the author just sets mu to epsilon in the second term, giving epsilon^2 in the denominator, then subtracts all the correction terms from 1 on the LHS and takes the 2/3 root to isolate mu/epsilon. Unable to simplify these two equations. I mean, say, as the result of some calculation I get a symbolic equation a x^2 + b x + c y, and say I am interested in whatever is multiplied by x^2 (i. 3. I have 2 equations that I want to equate to each other and solve for a certain variable in one of the equations. I don't know any way to solve this problem completely automatically in Mathematica, but it can definitely help in semi-automatic solution. Returning an equation from a function. In general, that's not possible - that is, for an arbitrary expression subject to an arbitrary constraint, there's no guarantee that the To simplify equations, combine like terms, To solve a linear equation, get the variable on one side of the equation by using inverse operations. Hot Network Questions Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Assign the results from a Solve to variable(s) In the following example: Clear[g1, z1, a, g2, z2] sa = Solve[ g1 == 1 + l z1/(z1^2 + a), a] sb = Solve[ g2 == 1 - l z2/(z2^2 + b), b] /. I understand that Mathematica evaluates symbols as early as possible, and so I have tried playing with Hold but it didn't help. But I want to limit the domain to 0 or 1 in this equation 2 x11 + 8 x12 + 6 x13 - 4 x14 + 7 x15 + 3 x16 - 5 x17 + 4 x18 + 2 x19 + 2 x20 + 10 x21 + $\begingroup$ Apologies if this seems pedantic, but I'm not sure I understand the question. I'm looking for some way to put the set of equations in to a variable and refer to it latter in NDSolve to make the code more easy to read. Using the output of Solve. Here the input (in FullForm) How to solve equation in Let's say that I need to know all the solutions to an equation like (variables in an interval) c1 = 1 <= x <= 9; c2 = 1 <= y <= 9; c3 = 1 < = z <= 9; c4 Solve equation with conditions of a variable equal to a set of values. In other words, we want to move everything except "x" (or whatever name the variable has) Unless I'm mistaken, the reason why this doesn't work is that Solve and Reduce do not have an Assumptions option, so Assuming has no effect on them. f (flavors). how to solve equation over positive integers in To solve the same equation for b, I need to write: Solve[{a*x + b*y == c}, {b}] It will give me: b -> (c - a*x)/y Similarly, I will have to repeat the steps for all variables like x and y. As shown by @Nasser, the four equations are obtained by equating the corresponding coefficients. ; If not enough end conditions are specified, RSolve will give general solutions in which undetermined constants are introduced. Solve[p == 2 t + 1, t][[1]] Mostly I don't like to use the different variable name within Solve. Attempting with the following input: Solve[{x == (1/2) y}, y] Yields the result output: {{y -> 0}} I'm using NDSolve in Mathematica to solve a set of equations. I'm trying to figure out how to use Mathematica to solve systems of equations where Suppose I want to solve a simple equation in mathematica x-a=0. My presenting case is an enormous reduced equation in many variables that contains (say) a single a amidst many terms Since you have only 5 equations, you can not eliminate all six a[] to get one equation with all b's. Sometimes the free package Guess. Thanks I would like to understand the reasons and find a way to avoid such behaviour of the Solve function in Mathematica 8. I've also tried to recast the problem as simultaneous equations and then using Solve or Eliminate as mentioned here: Rewriting expression in terms Obviously the equation cannot be solved (1 equation in several variables). Off the top of my head, the key differences are as follows: 1) Reduce simplifies logical statements, while Solve solves equations. The equations are written in the form of lefthandside == righthandside. Mathematica. I use the following construction for acquiring one solution from the expression returned by Solve. make a Variable with Subscript in Mathematica. I think a solution that is more elegant must The system will then be solved with a pseudo-spectral method for spatial discretization and a Runge-Kutta method for time integration, in the spirit of this question, but I'm having trouble with defining the new equation and the matrix $\hat{L}$ under the change of variables, in order to use it in the latter methods, can someone offer a proposal? Consider the operation Solve[f[x,y,z] == 0, x] I want to put the solutions for x obtained by this operation as the new variable. Commented Mar 6, Solving a system with 4 variables and 4 equations. My variables are a1,a2, a1prime, a2prime. I would appreciate if someone could tell me what I'm doing wrong. Thanks for contributing an answer to Mathematica Stack Exchange! Solve::svars: Equations may not give solutions for all "solve" variables. For example, the equation Well, I need a way to solve equations by another var got from other equation in Mathematica 8. For example: Solve[x^3 + 1 == 0, x] That syntax makes Mathematica assume that y can be anything, and if y is different from 13/2, then there's no x, so it can't solve it for the general case. We know that, in general, an equation system defines a set of points as a function of the parameters in it. Normally, I would do this with basic algebra and identities. there are many many complex solutions so the restriction of the domain to Reals is important to get Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. Express in terms of a new variable. For example, x^2+y^2 /. If no variables are specified, the result is a logical combination of terms fHx 1, , x n Lã0 and gHx 1, , x n L≠0, where f and g are polynomials, and each x i is a free variable of the system. Why can mathematica not solve my equation? 0. Is there an equivalent for strictly positive Reals? Something like PosReals I would like to Solve an equation so that the solution is in terms of multiple variables. Then, name the first equation f for convenience, Divide one side of the equation by the other, take the Log of the result, FullSimplify, and divide by the Log term. My idea was something like this: For[i = 0, i < 4, i++, For[j = 5, j < 10, j++, Print[i + j]]] Except that print, of course, is the Solve[expr, vars] attempts to solve the system expr of equations or inequalities for the variables vars. I am having trouble moving certain variables from one side of the equation to the other. Share. Can someone give me a little nudge in the right direction? How do I tell mathematica that I want to solve for v0? $\begingroup$ The first example under Basic Examples should show you all you need. I had another equation, z=x+2+4, I would like to know if there is a way in Mathematica to get an expression in which z= a*y+C, in which a and C are combinations of constants. I cannot get an analytic solution (not surprised), but NSolve is not doing the job for me. Get help on the web or with our math app. 3 How to solve an equation with I'm trying to figure out how to use Mathematica to solve systems of equations where some of the variables and coefficients are vectors. Why can mathematica not solve my equation? 1. Hot Network Questions Are qualia an illusion? Find the UK ceremonial county of a lat/long pair There is no single command that accomplishes the simplification desired. But note that this will change the definition region. Mathematica can solve some two-variable recurrence equations, but not all. I've tried the following Reduce[-2 (m^2/36 + n^2/4)^(1/4) Sin[1/2 When you use Solve, you are finding values for some of the variables in terms of others, subject to the constraints represented by the equations. Solve takes two arguments, the first is the equation to be used and the second is the variable that the equation is to solved in terms of. Solve[x + y + z == 5 && y == 3 , {x, y You would then instead formulate the optimization problem in terms of a merit function that you Why won't Mathematica Solve a set of two equations for one variable? 4. which is true. 5. syms g0 g1 g2 g3 x mu3 mu4 mu5 mu6 gamma A = Mathematics Meta matlab symbolic solve system of linear equations in terms of specific variable. Common choices of dom are Reals, Integers, and Complexes. In Mathematica if I have a symbolic algebra equation, how can I get the factor which is multiplied by a specific term?. (You must request access to the file; contact information is on that page. Related questions. I'm trying to use Mathematica to solve a set of equations. For example, DSolve: There are more dependent variables than equations, so the system is underdetermined. How to solve equation in terms of the variable. I'd like to filter the results of a solve giving inequality conditions: let's say I have Solve[f[x]==g[y],y] look at the documentation page for Solve where states clearly that inequalities can be directly given as part of the equations. Commented Mar 14, 2020 at 21:59. I'm looking for insight into other techniques that might help me solve the set of equations other then Solve. >> Solve::ratnz: "Solve was unable to solve the system with inexact coefficients. How to find exact value of a variable In Mathematica you need to make this relationship explicit by using y[x] in place of y. Suppose there is an equation like this (overly simplified example): eqn1 = ui - ua r1 / (r1 + r2) == 0 How can I get Mathematica to "solve" the equation in terms of: Keeping equations in terms of specific variable. a x^(17/6) - b x^2 - 1 == 0 where a and b are variable coefficients (which are very small and positive). Note that Solve may not be able to find expressions in terms of these variables (if the equations you give it are contradictory or insufficient) or for all values as some functions have no inverse or only partial inverses. I am trying to find an analytical solution to these equations, if it exists, using Mathematica. $\endgroup$ (Having three equations and four unknowns, it is most likely that any three of the variables can be expressed in terms of the fourth, or alternately that any one variable can be expressed solely in terms of each of the other three variables. Understanding solution to differential equation in Mathematica. All are solved with the skipped variable as linear factor: With other words, the missing condition fixes the up-to-now free linear factor. How can I simplify an expression in terms of a specific variable? To simplify an expression in terms of a specific I am trying to solve an equation by assuming that all the variables are real and strictly positive. It's often useful to get just one number from Solve. More than just an online equation solver. To learn more, Since version 8, Solve and Reduce share a great deal of code. 🌐 Languages: EN, ES, PT & more: 🏆 Practice: Improve your math skills: 😍 Step by How can I write a code in Mathematica to solve an equation with variable in specific domains? For instance, I want to write a code for Solve[{2 x + 1 == y && x > 0 && y > 0}, Basically, though: you probably want Eliminate, which eliminates variables from systems of equations. Mathematica: 'Findroot' not returning expected result. Wolfram Mathematica solving a particular equation. " Second, Solve may use nonequivalent transformations. I used Solve, but it didn't work. The definitive Wolfram Language and notebook experience. g. Use MathJax to format equations. Rule -> Set. Using the Mathematica - Solve equation as an expression of two variables. I basically want to solve these simyltaneous equiations: NSolve[ { - Yes, you can use multiple variables in the ReplaceAll function to manipulate an expression. Provide details and share your research! But avoid . Solve equation using Mathematica. (the short form of ReplaceAll) to apply the rule: Solve[expr, vars] attempts to solve the system expr of equations or inequalities for the variables vars. Separation of variables with Mathematica Solve the equation dy••••• dx =-xI1+y2M with initial conditions y(2)=1. 37037\exp u-u}{u-1}\mathrm du . For example to rearrange . Wolfram|One. Let's start with two vector equations since they can be easily written as equation for arbitrary vector component: You only have one equation to solve a simultaneous equation. So I don't think it's been solved correctly. The original First, write the coefficient matrix , variable vector and constant vector : Verify the rewrite: LinearSolve gives a Solve a million equations using an iterative Solve an Equation with Constants on Both Sides. Mathematics Equation Solving Wolfram Language Symbolic Computations. When there is no solution, Solve returns {} and the post-processing replacement rule {} -> {#} simply replaces the no-solution {} with the current solution. using ==), then use Solve to solve for expr specifying that the R and a variables should be eliminated: Clear Thanks for contributing an answer to Mathematica Stack Exchange! Keeping equations in terms of specific variable. I am not sure about the details of your studies and hence cannot First of all, you have given the output for specific values of other variables, which you have not shown us. , In[1]:= g = Solve getting solution to an equation as a variable with subscripts. How can I get mathematica to solve the non-homogeneous differential equation with undetermined How can I get Mathematica to solve an equation with multiple variables? Ask Question Asked 12 years, 10 months ago. Rewrite your definitions as equations (i. equation $\endgroup$ 1 $\begingroup$ You have a complicated I would like to solve the system A*B=I for the elements of B in terms of the elements of A. In general, you can not write t in terms of only t: p[t]. The output I am looking for is y == 2x. Quadratic Equation Solver PHP. However, x should also change after a certain number of runs of y. (1/a)-(b*e^-2a)/a= -(e^-2a*V0)/a I want to find a general expression for V0 where a and b are constants. So I am writing the mathematica code for this as below: Solve[x - a == 0, x] So the output will be as below: {{x -> a}} Now suppose I have assigned a value for 'a' beforehand and then want to solve the same equation. I am a new user to Mathematica, and would like to accomplish a very straightforward and simple goal: Given the equation x == (1/2)y, I would like to rewrite in terms of y. This is separable with gHxL=-x and hHyL=•••••1•••••• 1+y2. Actually I am trying to solve for several n's equations self consistently and get the values for z where I need to have convergence after certain value of n. The ConditionalExpression output by Solve in your case does not really depend on the whole argument of the Log function in your original equation, but only on the following expression:-π < Im[(k t (L - T))/L] ≤ π If it is possible in your case to make assumptions on the values of those parameters, you could then try to Simplify the output of Solve using the LinearSolve[a, b] finds an x that solves the array equation a . If there were a way to force it to solve for K in terms of p,a,b,c,d, this might work, How to perform a complicated change of variables for a polynomial (in Mathematica) 7. Ask Question Asked 2 years ago. >> And the output just gives B in terms of A and t. In other words, Solve must return results that satisfy the equalities. Note the underscores; they matter (in a way that will be explained later). Your data are your list of values, the instead of Log[a+b x^a] put the form of your model, instead of {a,b}, your have three fit parameters {a,b,c} and you One can Fold Solve into the equations and thread the solutions generated at each step. Can't get solutions in output of wolfram mathematica. Products. Thank you! Below is my equation. Making statements based on opinion; back them up with references or personal experience. $\endgroup$ – MikeY. NDSolve::derivs: No derivatives of dependent variables were found in the equations. Are you just asking how to get from 7. And usually that's how we draw phase diagrams, that demonstrate conformal windows, $\chi SB$ phases, and so on. How do i change variables of function in mathematica ? for example of f(p,q) = p^3 You can use Solve to find the old variables in terms of the new variables, Use MathJax to format equations. Wolfram|Alpha is a great tool for finding polynomial roots and solving systems of equations. NDSolve is designed to solve differential or differential algebraic equations. POSTED BY: Solve Equations from Equations Mathematica. $\endgroup$ – Moo. I hope You can help me what to do with this beauty in order to get a solution. I am trying to solve a linear system of four equations with Mathematica. If that is not acceptable, then the question needs to specify how the I have a system of second order differential equations, with two independent variables, In other words, your results are impacting plotting over the entire range. However, I certainly agree that you could bring this to thea attention of WRI, because it is clear that Eliminate can solve your problem, albeit with a I solved a set of equations with indexed variables using Solve. Can you restart the Kernel or Mathematica and try again? $\endgroup$ – halirutan. I have a task to solve equation system with FindRoot: a*x+b*y^2-c*x^2=a-b, Mathematicas: How to solve equation in terms of the variable. {-3t} + Sin(3t)$ Right now i'm focusing on the particular solution with the term $\ 2t^4 $ hence the above Solve::svars: Equations may not give solutions for all "solve" variables. One can go in by hand and identify the desired term and manipulate the equation, basically by hand but that is of little help. Mathematica doesn't allow implicit definition of a matrix, so your Z should be defined as something like:. >> How to ask Mathematica to express variables in terms of each other in an implicit function. Equation simplification issue in Mathematica. Now I want to have these indexed variables store their values. Using Reduce. At least one of the two parameters a and e must assume a special value for the solution set not to be empty", this can be done as follows. In these equations there are four real and positive parameters a1, a2, m, k. I can use the keyword Reals for the argument dom in the Solve function Solve[expr,vars,dom]. Equations are given in the form lhs == rhs. 'Assuming[alpha>beta>0,Solve[Cos(alpha*Cos(x)) + Cos(beta*Cos(x)) -1. Equations such as a [0] == val can be given to specify end conditions. Is there a way to tell Solve (or another function) that x2,x3,x4 are not to appear in the solution for x1, but that there is no need to solve for them? If so, please, how? Indeed, I would much rather that Solve return the coefficients of the polynomial for x1 , rather than returning the solution to that (perhaps high-order) polynomial. Is there a more specific technique I could use that's more optimized for this kind of problem specifically? Leaving out ForAll yields solutions in terms of the x-variables, which is kind of useless. An approximate answer for each variable is acceptable. f (colors) or global d. Second-order equation with variable coefficients, solved in terms of elementary functions: Airy's equation: Fourth-order equation solved in terms of Kelvin functions: Linear equation with q-rational coefficients: Higher-order inhomogeneous equation with constant coefficients: Online math solver with free step by step solutions to algebra, calculus, and other math problems. Asking for help, clarification, or responding to other answers. Evidentally, the way you are coding introduces some loss of transparency. Ask Question Asked 7 years ago. Of course the "help" file is no help For obvious and analogous reasons, Gather does not solve this problem either. z2 -> z1 + l This gives me solutions for a and b that are in the form {{ a -> astuff }} {{ b -> bstuff }} I want to use the linear version of a somewhat big equation which is outputted by my Mathematica code - For simplicity I will here use the example equation: Test = 3 x + x y + 8 y Now, I want to use only the first order term, so that for x that will be 3 and for y that will be 8. e. I am using Mathematica. This particular problem I can solve on paper, but others are much more complex. Commented Aug 2, 2016 at 11:51 $\begingroup$ Just to confirm: you want to solve $$1=x^2+\int_{1. In fact, by Specifying Method -> Reduce in Solve, Solve will use Reduce behind the scenes to produce an answer. How can I do it? Mathematica is often quite different to other programming languages I can use. $\begingroup$ You have 2 functions of 3 variables: t[a,b,c] and p[a,b,c]. Mathematica - Solve equation as an expression of two variables. Example: x − 2 = 4. For them to be true, you need to solve for y also Here are some equations for illustration: Format[Solve[{4*r1^3 == r2^2*(3*h), t*r2^2*(h - d) == r2^2*(h + d Is there any way to force Mathematica to output the results as: {{{h -> (4*r1^3)/(3*r2^2), d -> h*(-1 Figuring out a way to express one variable in terms of another will simplify them a lot. However, what you may try to do is to write p in terms of e. a in here). But Reduce with the command to forced elimination of all a's, gives you at least a few relations for b's. Its syntax is Solve [eqns, vars], where eqns is your equation or set of equations and vars are the variable (s) in the The Wolfram Language's symbolic architecture allows both equations and their solutions to be conveniently given in symbolic form, and immediately integrated into computations and To put an expression in terms of a specific variable in Mathematica, you can use the ReplaceAll function (symbol: /. Mathematicas: How to solve equation in terms of the variable. I started with Solve[{y^x >= 2*y - 1}, x] but Use MathJax to format equations. You want to solve for r but not in terms of ts, which means you need to treat ts as additional variables for solving. For example, if i want to solve for x1: Reduce[x1 + x2 + x3 == 4, x1] x1 == 4 - x2 - x3 but if i want to solve for 2 variables (x1 and x2): Reduce[x1 + x2 + x3 == 4, {x1, x2}] Is it possible to have Mathematica move all terms to one side of I want to try to solve the following . Ask Question Solve and Reduce do not work in this case. You may have noticed that in all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. Third, use Reduce instead of Solve. Define function in Mathematica with previous output. 0. Is there an easier to tell Mathematica to solve an equation for all variables one by one? Thanks. Start by defining Mathematica functions for g and h. Is there any way how to ask Wolfram Alpha to "redefine" (not solve) equation so I can see something like following: I tried to do it manually (but result is not OK) b=((P/EY)*12A))/t^3 I wish to see how right equation will look. I want to solve an equation for one value x and many values y. The Wolfram Language's symbolic architecture allows both equations and their solutions to be conveniently given in symbolic I'm not sure how to get the results in terms of the desired variables. Variables can be any expressions. In Solve an equation with specific domain? I asked how to write a code in Mathematica with one specific domain for one of its variables. Reduce[a + I b == zr + I zi && Element[{a, b, zr, zi}, Reals], zi] (* ==> (zr | b) \[Element] Reals && a == zr && zi == b *) $\begingroup$ If you are very certain that you obtain only one set of solutions from Solve, then you could use Solve[youreqautions, yourvariables] /. NSolve [expr, vars] assumes by default that quantities appearing algebraically in inequalities are real, while all other quantities are complex. Solve[expr, vars, dom] solves over the domain dom. $\endgroup$ – I'm trying to use Mathematica to solve for t in this equation: Sum[Subscript[a, i]E^(-Subscript[g, i]*t), {i, 0, getting solution to an equation as a variable with subscripts. and i know every variables but not "b". ) Wolfram Community forum discussion about Avoid message "Equations may not give solutions for all "solve" variables"?. Not surprisingly, this doesn't work. Z = Table[z[i, j], {i, 2}, {j, 2}]; I am trying to solve a system of equations (5 unknown variables, 5 equations) but the Solve[] function just hangs and I have to abort the evaluation. With variables specified in the input, Resolve gives the same while Mathematica 9 raises. where. These equations are examples, the equations I working are more complicated (thanks for the patience I am newbie $\begingroup$ I agree that often you just want to use rules, but in my case I'm trying to set a global variable before going on to solve for other things (where Solve will work better if it has this global value instead of trying to remain general for all possible values). By using Solve equations can be rearranged and put in terms of different variables within them. The data are experimental data and won't produce exact perfect answers for every equation. The original technical computing environment. equation in [,] I simplified by A Get the free "Solve for Variables" widget for your website, blog, Wordpress, Blogger, Added Aug 1, 2010 by ThorsSon in Mathematics. Free solve for a variable calculator - solve the equation for different variables step-by-step When a single variable is specified and a particular root of an equation has multiplicity greater than one, NSolve gives several copies of the corresponding solution. A Solution is a value we can put in place of a variable (such as x) that makes the equation true. Learn how to solve an equation with plenty of examples. Solve Equations from Equations Mathematica. I tried the equivalent of Set[sols[[1]]], and when that didn't work. o. 1. Here is a simpler version of what I'm doing: In[1]:= equs = {a x + b y == 0, x - y == 1}; In[2]:= f[a_, b_] := Module[{x, y}, {x, y} /. SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. Basically I want to solve the differential equation for alpha, which I suppose is actually a function of four variables. We can tell Reduce that these variables are all real-valued like this:. It also factors polynomials, plots polynomial solution sets and inequalities and more. Why users love our Equation Calculator. But I'm guessing that what you meant to ask was how you can find the value of an expression (a+b) subject to a constraining equation (a^3 + 3 a^2 b + 3 a b^2 + b^3 == c). Eliminate works primarily with linear and polynomial equations. I am having trouble getting Mathematica to solve really simple inequalities. ) It lets you solve an equation for some constant or for some variable within the equation. I will give an elaborate answer because I think this is an area where Mathematica really shines ;) Let me digress. ; In NSolve [expr, vars, Reals] all variables, parameters, constants, and function Built into the Wolfram Language is the world's largest collection of both numerical and symbolic equation solving capabilities\[LongDash]with many original algorithms, all automatically accessed through a small number of exceptionally powerful functions. This does not happen all the time—so now we’ll see how to solve equations where the variable terms and/or constant terms are on both sides of the I would like to solve the following system of equation in terms of g3 The following code returns one possible solution. A Linear Solve Mathematica. You can also click "show steps" in the results to see how it got there. t,b,c: p[t,b,c]. Solve an equation with two unknowns, a and b. Common choices of dom are Reals, It would seem (at least to me) that a perfectly valid answer to Solve would be $-a_{2x}$ but that would be a useless answer, as I want the answer in terms of $m_1 , m_2 , g, \theta$ and not in just use the Solve function. MathJax reference. Z = {{z1, z2}, {z3, z4}}; For the case in which the order of Z is very high, you may prefer this kind of definition (just in case, I'd like to mention that something like Z(1, 1) doesn't make sense in Mathematica):. Thanks for contributing an answer to Mathematica Stack Exchange! Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site This yields the correct solution on a smaller equation (about 20 variables and 9 parameters) but is far too slow for this one. Sign Isolate variable in inequality. However, a string of commands can accomplish it. In[1]:=g@x_D Solve is the Mathematica function used for symbolically solving a polynomial equation or set of equations. When we put 6 in place of x we get: 6 − 2 = 4. X and y shall of course have some kind of max value for which it shall evaluate the solve function. I have a beautiful equation, where I am trying to compute for R2. Note that once those variables have values assigned, your Solve expression will no longer work until the values are I have a system of equations for algebraic curve given by the zero locus of some polynomial encoded in the system of equations (I want to eliminate variable z and get algebraic curve in terms of x and y, C12 and C22 can be assumed to Confirming that Mathematica 9 can easily solve this system unlike ver. I want to get an expression of x in terms of unknown constants alpha and beta. Input an equation with as many variables as you want. How can I get the output solution to be stored in the variables themselves? I obtained the results from Solve in the form: f[1,1,1]->3, f[1,1,2]->5,. ). 2. 96 ==0,x]] ' Built into the Wolfram Language is the world' s largest collection of both numerical and symbolic equation solving capabilities with many original algorithms There are four major areas in the study of ordinary differential equations that are of interest in pure and applied science. m can solve what RSolve cannot. However, I am still trying to solve all of these two errors and trying to get the expression of this function in terms of one variable. of course, C[1] is an arbitrary constant, so making transformations involving it is a nicety. Namely, the set of variable assignments or I want to solve the following equation. {x->a, y->b} will transform the expression x^2+y^2 into a^2+b^2. A single variable or a list of variables can be specified. The specification a ∈ Vectors [m] or a ∈ Matrices [{m, p}] can be used to indicate that the dependent variable a is a vector-valued or a matrix-valued variable, respectively. This is done automatically in the script by replacing all occurances of y with y[x]. I have an equation that defines a variable like y=4x+a. However, is it possible to use Mathematica When using the Solve function with Mathematica, you can specify for what variables you want Solve to specify the solutions. . Resolve can eliminate quantifiers from arbitrary complex polynomial systems. Provide details and share your research! But avoid Asking for help, clarification, or responding to other answers. Then specify the variable you want the equation to be solved in terms of. So far every single time that I have tried to use Solve[] or Reduce [] the computation takes forever. Of these four areas, the study of exact solutions has the longest history, dating back to the period just after the I'm using Wolfram Mathematica 8 and I want to make it solve this equation for the variable t: I have tried to type your equation in Mathematica 8 and it was solved. Enter your queries using plain English. $\begingroup$ As I performed simplifications of the left side of the equation inside Solve, I did the same on the right side, namely C[1]. Solve a linear system of equations in R. 66? From 7. Now I have a questions: If two of the variables are from specific domain for which they have been taken correspondingly? for example: Solve[{z == 2 x + y && x > 0 && y > 0}, {x, y, z}, Integers] // Column I have a function f(x,t) and I'd like to plot the function of the solution x(t) of f(x(t),t)=0 using Mathematica. My [ , ] with one of your complicated terms. And the rule-applying seems very unnatural. Simultaneous equations can be combined either in a list or with &&. Replacing x by 1/2, the solution you're looking for, doesn't make the equalities be true. 4. How to properly attempt to express a function in terms of new variables. Last, but not least: your code is irreproducible because you use RandomReal without Block. (Solve automatically choose which variable to solve for. I was thinking of a nested for-loop, as I am still relatively new to Mathematica. So my code will look like below: a = 1; Solve[x - a == 0, x] $\begingroup$ First, Solve also produces a warning "Solve::svars: Equations may not give solutions for all "solve" variables. Normally, I would try something looking like: Create arrays X, T For t in T do solve (numerically) f(x,t)=0, append the solution to X Plot X Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Basically there are four variables, where a,b,c are constants related to the physical apparatus which I'll put in later, and then alpha, which is the main variable I want to consider. I can not get my head around why mathematica can not solve this equation: In[22]:= Solve[1/x^12 - 2/x^6 + 1/2 (-2 + x)^2 HeavisideTheta[-2 + x] == 0] During evaluation of In[22]:= Solve::nsmet: This system cannot be solved $\begingroup$ @C. To learn more, see our tips on writing great answers. q = t /. The answer was obtained by solving a corresponding exact When you solve equations with multiple variables using solve, the order in which you specify the variables can affect the solutions. = is used to set variables; == is used to test equality. How to solve an equation system in Mathematica 9. I would always have the same number of equations and variables you are solving for. Does anyone know how to solve this in Matlab or Mathematica? Thanks. fdaiay heagt uibcin uaxoo hfxuo bxbw pkdcb uxeaqo htsq kksyx