3
$\begingroup$

How to solve the following integral equation in Mathematica?

eqn = y[x, z] == z + x*Integrate[y[t, v], {t, 0, 1}, {v, 0, 1}]
sol = DSolveValue[eqn, y[x, z], {x, z}]

What is the problem? Mathematica usually gives me answers like that while solving things that look difficult. I can solve simple problems. Difficult problems are why I need a software. Thank you in advance.

$\endgroup$
5
  • 2
    $\begingroup$ Please post your code, not a picture of it, so that readers can try to help you without having to transcribe it. Also, do yo9u have reason to believe that a symbolic solution actually exist? $\endgroup$
    – bbgodfrey
    Oct 22 at 22:27
  • $\begingroup$ @bbgodfrey I do not believe. I am seeing that it is not working. If it doesn't exist, why doesn't it give an error? or say "it doesn't exist"? If I knew the answer I wouldn't try in the first place. $\endgroup$
    – user82393
    Oct 22 at 22:54
  • $\begingroup$ In fact, Mathematica almost always returns unevaluated, when it cannot solve a problem. I agree that it should provide an explanation, but it does not. By the way, a solution to your second equation does exist, even thought Mathematica cannot find it. It is x + z . $\endgroup$
    – bbgodfrey
    Oct 22 at 23:16
  • $\begingroup$ @bbgodfrey Thank you so much. The actual equation I am trying to solve is more complicated than this but it has the same shape. I would be pleased if you could recommend me another software/application. $\endgroup$
    – user82393
    Oct 23 at 9:26
  • $\begingroup$ Please post your code instead of picture. $\endgroup$
    – cvgmt
    Oct 23 at 10:13
8
$\begingroup$

Mathematica can solve your equation, but it requires some understanding of mathematics and human intervention. This is always the case for nontrivial problems, isn't it?

Additive solution was demonstrated by @bbgodfrey. But there is another multiplicative one. Separate variables as follows

$$y(x,z)=g(x) h(z),$$ and set $$G=\int_0^1 g(x) dx.$$

Now your equation reads after integrating it $\int_0^1 dx\ldots$ $$ G h(z)=z+\frac{G}{2} \int_0^1\! h(v)\, dv. $$

This equation can be solved with MA

eqn = G* h[z] == z + 1/2* G*Integrate[h[v], {v, 0, 1}]
DSolveValue[eqn, h[z], z]

$$h(z)=\frac{2 z+1}{2 G}.$$

The full solution therefore reads

$$y(x,z)=\frac{(2 z+1)g(x)}{2 G},$$ where $g(x)$ is arbitrary function on the $[0,1]$ interval such that $G=\int_0^1 g(x) dx\neq0$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy