Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This equation is simple to DSolve in Mathematica, but I don't know how to use and understand solution.

 DSolve[Y1''[x] - a2 Y1[x] - a1*X1[x] == 0, Y1[x], x]

$ \text{Y1}[x]\to e^{\sqrt{\text{a2}} x} C[1]+e^{-\sqrt{\text{a2}} x} C[2]+e^{-\sqrt{\text{a2}} x} \left(e^{2 \sqrt{\text{a2}} x} \int_1^x \frac{\text{a1} e^{-\sqrt{\text{a2}} K[1]} \text{X1}[K[1]]}{2 \sqrt{\text{a2}}} \, dK[1]+\int_1^x -\frac{\text{a1} e^{\sqrt{\text{a2}} K[2]} \text{X1}[K[2]]}{2 \sqrt{\text{a2}}} \, dK[2]\right) $

What is dK1 and dK2, how to solve integrals [1,x], and how to use this solution Y1[x] to find X1[x] in another equation

Derivative[2][X1][x] == Y1[x]
share|improve this question
up vote 3 down vote accepted

the dK[2] just means the integral is with respect to K[2],

$\int f(t) dt$ is exactly the same as $\int f(K[2]) dK[2]$

In this case, it appears in the solution because X1[x] is an unknown, so the only way for the differential equation to be formally solved is to include how it would affect the solution via the integrals. However, you can solve for x1 at the same time:

sol = DSolve[{
   y1''[x] - a2 y1[x] - a1 x1[x] == 0,
   x1''[x] == y1[x]},
  {y1[x], x1[x]}, x]

Since no initial conditions are supplied the result has a bunch of arbitrary constants called C[i] in this case there are four of them.

Here's an example using the result to create two functions f,g that correspond to y1,x1 with some specific constants:

Clear[f, g]
params = {a1 -> 1., a2 -> -2., C[1] -> 3., C[2] -> -3., C[3] -> 2., C[4] -> 3.};
f[x_] = y1[x] /. sol /. params;
g[x_] = x1[x] /. sol /. params;
Plot[{f[x], g[x]}, {x, 0, 1}]


share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.