# Why is Mathematica unable to solve this integral equation?

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.

• 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? Oct 22 at 22:27
• @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.
– user82393
Oct 22 at 22:54
• 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 . Oct 22 at 23:16
• @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.
– user82393
Oct 23 at 9:26

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]


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

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