Question

verify that the given differential equation is not exact. Multiply the given differential equation by the indicated integrating factor and verify that the new equation is exact. Solve.

$\left(x^2+2xy-y^2\right)\mathit{d}\mathit{x}\mathbf{+}\left(y^2+2xy-x^2\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{\mu }\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}}^{\mathbf{-}\mathbf{2}}$

Short answer

The required solution is$-x+y+\frac{2x^2}{x+y}=c$

Step-by-step solution

To Find whether the given differential equation is exact or not

The given equation is $\left(x^2+2xy-y^2\right)dx+\left(y^2+2xy-x^2\right)dy=0$.

Compare the given equation with$Mdx+Ndy=0$

$\begin{array}{l}M=\left(x^2+2xy-y^2\right)\\ N=\left(y^2+2xy-x^2\right)\end{array}$

As,$\frac{\partial M(x,y)}{\partial y}\ne \frac{\partial N(x,y)}{\partial x}$ . The equation is not exact.

The given equation can also be written as,

$\left({(x+y)}^2-2y^2\right)dx+\left({(x+y)}^2-2x^2\right)dy=0$

Then multiply both sides in integrating factor ${\left(x+y\right)}^{-2}=\frac{1}{{\left(x+y\right)}^2}$

$\begin{array}{c}\frac{{(x+y)}^2-2y^2}{{(x+y)}^2}dx+\frac{{(x+y)}^2-2x^2}{{(x+y)}^2}dy=0\\ \left(1-\frac{2y^2}{{(x+y)}^2}\right)dx+\left(1-\frac{2x^2}{{(x+y)}^2}\right)dy=0\end{array}$

From the obtained equation, it can be observed that,

$\begin{array}{c}M(x,y)=1-\frac{2y^2}{{(x+y)}^2}\\ N\left(x,y\right)=1-\frac{2x^2}{{\left(x+y\right)}^2}\end{array}$

$\begin{array}{l}\frac{\partial M}{\partial y}=-\frac{4y{(x+y)}^2-2y^2\times 2(x+y)}{{(x+y)}^4}\\ \frac{\partial M}{\partial y}=-\frac{4x^{2}y+4x{y}^2}{{(x+y)}^4}\\ \frac{\partial N}{\partial x}=-\frac{4x{(x+y)}^2-2x^2\cdot 2(x+y)}{{(x+y)}^4}\\ \frac{\partial N}{\partial x}=-\frac{4x^{2}y+4x{y}^2}{{(x+y)}^4}\end{array}$

As,$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

Therefore, the equation is exact.

Find the solution

Then we have

$\begin{array}{l}g\left(x\right)=-x\\ g^{¢}\left(x\right)=-\frac{x^2+2xy+y^2}{{(x+y)}^2}=-1\end{array}$

$f(x,y)=y+\frac{2x^2}{x+y}+g\left(x\right)IntegrateN(x,y)=\frac{\partial f}{\partial y}=1\frac{2x^2}{{(x+y)}^2}$

Integrate$N(x,y)=\frac{\partial f}{\partial y}=1\frac{2x^2}{{(x+y)}^2}$

$\begin{array}{l}-x+y+\frac{2x^2}{x+y}=c\\ \frac{\partial f}{\partial x}=\frac{4x(x+y)-2x^2}{{(x+y)}^2}+g^{¢}\left(x\right)\end{array}$

Take $\begin{array}{l}\\ \frac{\partial f}{\partial x}\end{array}$

$\frac{\partial f}{\partial x}=\frac{2x^2+4xy}{{(x+y)}^2}+g^{¢}\left(x\right)g^{¢}\left(x\right)=-1$

Since, $M(x,y)=\frac{\partial f}{\partial x}$

$\begin{array}{c}g\left(x\right)=-x\\ g^{¢}\left(x\right)=-\frac{x^2+2xy+y^2}{{(x+y)}^2}\\ =-1\end{array}$

$Hence,therequiredsolutionis-x+y+\frac{2x^2}{x+y}=c.$