Answer to Question #108719 in Differential Equations for Aditi Yadav

Question

Solve the differential equation : (xy^2 - x^2) dx +(3x^2y^2 +x^2y-2x^3+y^2) dy =0.

Answer 1

Lets rewrite the equation as below:

$$\left(xy^2-x^2\right)dx+\left(3x^2y^2+x^2y-2x^3+y^2\right)dy=0\\[0.3cm] p(x,y)dx+q(x,y)dy=0\\[0.3cm] \left\{\begin{array}{l} p(x,y)=xy^2-x^2\\[0.3cm] q(x,y)=3x^2y^2+x^2y-2x^3+y^2 \end{array}\right.$$

The equation is called exact differential equation if it satisfied:

$$\frac{\partial p(x,y)}{\partial y}=\frac{\partial q(x,y)}{\partial y}$$

In our case,

$$\left\{\begin{array}{l} \displaystyle\frac{\partial p}{\partial y}=2xy\\[0.3cm] \displaystyle\frac{\partial q}{\partial x}=6xy^2+2xy-6x^2 \end{array}\right.\longrightarrow\\[0.3cm] \frac{\partial p}{\partial y}\neq\frac{\partial q}{\partial x}$$

Since the relation is not equal, this differential equation is not exact.

Next step is find a function $u(x,y)$ which will make equation 

$u(x,y)(p(x,y)dx+q(x,y)dy)=0$  become exact differential equation.

$u(x,y)$  is called integrating factor, and will be much easier to find if it is only $x$ function or only $y$ function.

$u(x,y)$  is only x function if

$$u(x,y)\equiv u(x)=\frac{\displaystyle\frac{\partial p}{\partial y}-\displaystyle\frac{\partial q}{\partial x}}{q(x,y)}$$

only have $x$ variable, and it is only $y$ function if

$$u(x,y)\equiv u(y)=\frac{\displaystyle\frac{\partial q}{\partial x}-\displaystyle\frac{\partial p}{\partial y}}{p(x,y)}$$

only have $y$ variable.

In our case,

$$\frac{\displaystyle\frac{\partial p}{\partial y}-\displaystyle\frac{\partial q}{\partial x}}{q(x,y)}=\frac{2xy-\left(6xy^2+2xy-6x^2\right)}{3x^2y^2+x^2y-2x^3+y^2}\\[0.3cm] \frac{\displaystyle\frac{\partial p}{\partial y}-\displaystyle\frac{\partial q}{\partial x}}{q(x,y)}=\frac{6x\left(x-y^2\right)}{3x^2y^2+x^2y-2x^3+y^2}\\[0.3cm]$$

contain both $x$ and $y$ .

$$\frac{\displaystyle\frac{\partial q}{\partial x}-\displaystyle\frac{\partial p}{\partial y}}{p(x,y)}=\frac{\left(6xy^2+2xy-6x^2\right)-2xy}{xy^2-x^2}\\[0.3cm] \frac{\displaystyle\frac{\partial q}{\partial x}-\displaystyle\frac{\partial p}{\partial y}}{p(x,y)}=\frac{6x\left(y^2-x\right)}{x(y^2-x)}=6\\[0.3cm]$$

contain only $y$ variable.

So $u(x,y)$  is only $y$ function. In this case:

$$u(y)=e^{\int 6dy}=e^{6y}$$

Back to our initial equation:

$$u(y)(p(x,y)dx+q(x,y)dy)=0\\[0.3cm] e^{6y}\left(xy^2-x^2\right)dx+e^{6y}\left(3x^2y^2+x^2y-2x^3+y^2\right)dy=0$$

Now find a function $f(x,y)$  that satisfied:

$$\left\{\begin{array}{l} \displaystyle\frac{\partial f}{\partial x}=e^{6y}\left(xy^2-x^2\right)\\[0.3cm] \displaystyle\frac{\partial f}{\partial y}=e^{6y}\left(3x^2y^2+x^2y-2x^3+y^2\right) \end{array}\right.\\[0.3cm] \frac{\partial f}{\partial x}=e^{6y}\left(xy^2-x^2\right)\longrightarrow\\[0.3cm] f(x,y)=\int\left(e^{6y}\left(xy^2-x^2\right)\right)dx+g(y)\longrightarrow\\[0.3cm] \boxed{f(x,y)=e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+g(y)}\\[0.3cm] \frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\left(e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+g(y)\right)=\\[0.3cm] =6\cdot e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+e^{6y}\left(x^2y\right)+\frac{dg}{dy}=\\[0.3cm] =e^{6y}\left(3x^2y^2-2x^3+x^2y\right)+\frac{dg}{dy}$$

Then,

$$e^{6y}\left(3x^2y^2-2x^3+x^2y\right)+\frac{dg}{dy}=e^{6y}\left(3x^2y^2+x^2y-2x^3+y^2\right)\\[0.3cm] \frac{dg}{dy}=e^{6y}\cdot y^2\longrightarrow g(y)=\int\left(e^{6y}y^2\right)dy\\[0.3cm] \left(\text{integration by parts}\right) : \\[0.3cm] g(y)=\int\left(e^{6y}y^2\right)dy=\frac{y^2e^{6y}}{6}-\int\frac{2e^{6y}}{6}dy=\\[0.3cm] =\frac{y^2e^{6y}}{6}-\frac{1}{3}\cdot\left(\frac{ye^{6y}}{6}-\int\frac{e^{6y}}{6}\right)=\\[0.3cm] =\frac{y^2e^{6y}}{6}-\frac{ye^{6y}}{18}-\frac{e^{6y}}{108}\equiv\frac{e^{6y}\left(18y^2-6y+1\right)}{108}$$

Conclusion,

$$\boxed{f(x,y)=f(x,y)=e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+\frac{e^{6y}\left(18y^2-6y+1\right)}{108}}$$

And finally our equation can be rewriten as:

$$\frac{\partial f(x,y)}{\partial x}\cdot dx+\frac{\partial f(x,y)}{\partial y}\cdot dy=df(x,y)=0\\[0.3cm] f(x,y)=C\\[0.3cm] \boxed{e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+\frac{e^{6y}\left(18y^2-6y+1\right)}{108}=C}$$

ANSWER

General solution

$$e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}+\frac{18y^2-6y+1}{108}\right)=C$$