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I have an exact differential equation, $$\displaystyle e^{-x^2y^2}\big(1-2x^2y^2-4xy^3\big)dx+e^{-x^2y^2}\big(2-2x^3y-4x^2y^2\big)dy=0$$

To solve it, I should evaluate one of these integrals at first:

$$\int e^{-x^2y^2}\big(1-2x^2y^2-4xy^3\big)dx\quad \text{or}\quad \int e^{-x^2y^2}\big(2-2x^3y-4x^2y^2\big)dy$$But I don't know how to do it. For example for the first one we have,

$$\int e^{-x^2y^2}\big(1-2x^2y^2-4xy^3\big)dx=(\int e^{-x^2y^2}dx)-2y^2(\int x^2e^{-x^2y^2}dx)-4y^3(\int xe^{-x^2y^2}dx)$$I think the first integral $(\int e^{-x^2y^2}dx)$ hasn't elementary solution (like $\int e^{x^2}dx$) for the second one IBP would works and last one can be evaluated by using $u=x^2$ substitution.

Etemon
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1 Answers1

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You have to find a function $F$ such that

${\partial F(x,y) \over \partial x}=e^{-x^2y^2}(1-2x^2y^2-4xy^3) \tag{i}$ and ${\partial F(x,y) \over \partial y}=e^{-x^2y^2}(2-2x^3y-4x^2y^2)\tag{ii}$.

Then, $F(x,y)=\int e^{-x^2y^2}(1-2x^2y^2-4xy^3) \partial x + \phi(y)$ assuming $F$ satisfies $\text{(i)}$

or, you can make $F(x,y)= \int e^{-x^2y^2}(2-2x^3y-4x^2y^2)\partial y + \omega(x)$ instead, assuming $F$ satisfies $\text{(ii)}$

If you go with the first, that is, you assume $F$ satisfies $\text{(i)}$, then calculate partial derivative of $F$ with respect to $y$ and then equate it with RHS of $\text{(ii)}$.

In your attempt, I don't understand why you are doing regular integration rather than partial integration.

$\textbf{Method of grouping}$

You can simply regroup the terms as you already know that this is an exact equation. It can be noticed that the derivative of power of $e$, i.e., $-x^2y^2$ is $(-2xy^2dx-2yx^2dy)$ which is missing an $x$ to match the terms in the given equation. So you might want to try $d(xe^{-x^2y^2})= xe^{-x^2y^2} (-2xy^2dx-2yx^2dy) +e^{-x^2y^2}dx$ which gives half the terms of your exact equation.

The remaining terms $e^{-x^2y^2}(-4xy^3dx-4x^2y^2dy) +e^{-x^2y^2}2dy$ can be regrouped as $2ye^{-x^2y^2}(-2xy^2dx-2yx^2dy)+2e^{-x^2y^2}dy=d(2ye^{-x^2y^2})$

Therefore the given exact equation can be written as $d(xe^{-x^2y^2})+d(2ye^{-x^2y^2})=0$

or, $d(e^{-x^2y^2}(x+2y))=d(c)$

Hence, a one-parameter family of solutions of the given differential equation is given by

$e^{-x^2y^2}(x+2y)=c$, where $c$ is an arbitrary constant.

$\textbf{Edit (continuing further the first method):}$

Let $F$ satisfies $\text{(i)}$, then

$\displaystyle \begin{array}.F(x,y)=\int e^{-x^2y^2}(1-2x^2y^2-4xy^3) \partial x + \phi(y)\\ =\left(\int \underset 1 e^{-x^2y^2}.\underset 21 \partial x- 2y^2\int x^2e^{-x^2y^2} \partial x\right) -4y^3 \int xe^{-x^2y^2}\partial x +\phi(y)\\ =\left(e^{-x^2y^2}\int 1 \partial x-\int \left(\left(\frac{\partial}{\partial x} (e^{-x^2y^2})\right) \int 1. \partial x\right) \partial x -2y^2\int x^2e^{-x^2y^2} \partial x\right) -4y^3 \int xe^{-x^2y^2} \partial x +\phi(y)\\ =\left(x e^{-x^2y^2} -\int \left(-y^22xe^{-x^2y^2}\right)x \partial x-2y^2\int x^2e^{-x^2y^2} \partial x\right) -4y^3 \int xe^{-x^2y^2} \partial x +\phi(y)\\ =\left(x e^{-x^2y^2} +\require{cancel} \cancel{2y^2\int x^2e^{-x^2y^2} \partial x}-\cancel{2y^2\int x^2e^{-x^2y^2} \partial x}\right) -4y^3 \int xe^{-x^2y^2} \partial x +\phi(y)\\ =xe^{-x^2y^2}-4y^3 \int xe^{-x^2y^2} \partial x +\phi(y)\\ \text{(Let $-x^2y^2=u \implies \frac{\partial u}{\partial x}=-2y^2x \implies \partial x= \frac{-1}{2y^2x} \partial u$)}\\ =x e^{-x^2y^2} - \frac{4y^3}{-2y^2} \int e^u \partial u +\phi(y)\\ =x e^{-x^2y^2} +2y e^{-x^2y^2}+\phi(y)\\ =(x+2y) e^{-x^2y^2}+\phi(y)\end{array}$

Now,

$\begin{array}.{\partial F(x,y) \over \partial y}= {\partial \over \partial y}(xe^{-x^2y^2})+ {\partial \over \partial y}(2y e^{-x^2y^2})+{\partial (\phi(y) \over \partial y}\\ =x e^{-x^2y^2} (-2yx^2) +2y(-2yx^2e^{-x^2y^2}) +2e^{-x^2y^2} + \frac{d \phi(y)}{dy}\\ =e^{-x^2y^2}(-2x^3y-4x^2y^2+2) + \frac{d \phi(y)}{dy}\end{array}$

Using $\text{(ii)}$ we get,

$e^{-x^2y^2}(-2x^3y-4x^2y^2+2) + \frac{d \phi(y)}{dy}=e^{-x^2y^2}(2-2x^3y-4x^2y^2)$

$\implies \frac{d \phi(y)}{dy}=0 \implies \phi(y)=c_0$

Thus, $F(x,y)=(x+2y) e^{-x^2y^2}+c_0$

Hence, a one-parameter family of solutions of the given differential equation is $F(x,y)=c_1$, which may be expressed as

$(x+2y) e^{-x^2y^2}=c$, where $c=c_1-c_0 $ is an arbitrary constant.

Aman Kushwaha
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  • Thanks for the answer. I understood the grouping method. But for the first method (integration), can you please elaborate on how to find $F(x)$ ? We have $F_y=e^{-x^2y^2}(2-2x^3y-4x^2y^2)$ hence $\frac{\partial}{\partial{y}}\left(\int e^{-x^2y^2}(1-2x^2y^2-4xy^3) \partial x\right) + \phi'(y)=e^{-x^2y^2}(2-2x^3y-4x^2y^2)$. I don't know how to calculate partial derivative of the integral. – Etemon Aug 09 '21 at 18:43
  • Sorry, I meant " how to find $F(x,y)$ " – Etemon Aug 09 '21 at 18:51
  • @Soheil first calculate the partial integral. I'm giving a brief to do that: when doing partial integration w.r.t $x$, treat $y$ as a constant in integrand and while partial integrating w.r.t $y$, treat $x$ as constant. You did the partial derivative of $\phi(y)$ correct. For partial derivative of $v(x,y)$ w.r.t $y$, treat $x$ in $v(x,y)$ as constant. – Aman Kushwaha Aug 09 '21 at 18:57
  • But my issue (as I mentioned in the question) is evaluating $\int e^{-x^2y^2}dx$. I also checked This post it seems there isn't elementary way to do it. – Etemon Aug 09 '21 at 19:00
  • @Soheil You don't need to completely evaluate $\int e^{-x^2y^2} \partial x$. Once you start evaluating this integral using "by parts", the other term will be cancelled. I've edited my answer, the integration method is complete. Take a look – Aman Kushwaha Aug 10 '21 at 07:15
  • Thanks! I got it. – Etemon Aug 10 '21 at 07:46