# Solve differential equations like x' = Ax; A is a matrix and x is a vector

Posted 8 years ago
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 Hello everyone,I am a new user who is trying to solve linear differential equations like x' = A x where A is a matrix of constants. I have not been able to successfully enter such an equation into wolframAlpha. I have tried things like x' = [{2,1},{3,4}]x  to no avail. At best, it tells me that the formatting is incorrect and solves for x = 2.I have look through the available examples, but none seem to use a matrix. How would you solve dx/dt = Ax?
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Posted 8 years ago
 a = {{1, 2}, {3, 4}};DSolve[{x1'[t], x2'[t]} == a.{x1[t], x2[t]}, {x1[t], x2[t]}, t] {{x1[t] -> ((3 E^(1/2 (5 - Sqrt[33]) t) +         Sqrt[33] E^(1/2 (5 - Sqrt[33]) t) -         3 E^(1/2 (5 + Sqrt[33]) t) +         Sqrt[33] E^(1/2 (5 + Sqrt[33]) t)) C[1])/(     2 Sqrt[33]) - ((-3 + Sqrt[33]) (3 + Sqrt[33]) (E^(        1/2 (5 - Sqrt[33]) t) - E^(1/2 (5 + Sqrt[33]) t)) C[2])/(     12 Sqrt[33]),    x2[t] -> -Sqrt[(3/       11)] (E^(1/2 (5 - Sqrt[33]) t) - E^(1/2 (5 + Sqrt[33]) t)) C[      1] + ((-3 E^(1/2 (5 - Sqrt[33]) t) +        Sqrt[33] E^(1/2 (5 - Sqrt[33]) t) +        3 E^(1/2 (5 + Sqrt[33]) t) +        Sqrt[33] E^(1/2 (5 + Sqrt[33]) t)) C[2])/(2 Sqrt[33])}}is the closest I can get to what you want to do, in Mathematica.
Posted 8 years ago
 Sometimes Wolfram|Alpha correctly understands Mathematica syntax and it is always a good idea to try that if if you can't get a solution otherwise.I have not been able to get Wolfram|Alpha to use Simplify to get a result half this size by factoring out E^(-1/2(-5+Sqrt[33])t).http://www.wolframalpha.com/input/?i=DSolve%5B%7Bx1%27%5Bt%5D%2C+x2%27%5Bt%5D%7D+%3D%3D+%7B%7B1%2C+2%7D%2C+%7B3%2C+4%7D%7D.%7Bx1%5Bt%5D%2C+x2%5Bt%5D%7D%2C+%7Bx1%5Bt%5D%2C+x2%5Bt%5D%7D%2C+t%5D