Question #108719
Solve the differential equation : (xy^2 - x^2) dx +(3x^2y^2 +x^2y-2x^3+y^2) dy =0.
1
Expert's answer
2020-04-10T17:14:29-0400

Lets rewrite the equation as below:


(xy2x2)dx+(3x2y2+x2y2x3+y2)dy=0p(x,y)dx+q(x,y)dy=0{p(x,y)=xy2x2q(x,y)=3x2y2+x2y2x3+y2\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:



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

In our case,



{py=2xyqx=6xy2+2xy6x2pyqx\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)u(x,y) which will make equation 

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

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

u(x,y)u(x,y)  is only x function if



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

only have xx variable, and it is only yy function if



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



only have yy variable.

In our case,



pyqxq(x,y)=2xy(6xy2+2xy6x2)3x2y2+x2y2x3+y2pyqxq(x,y)=6x(xy2)3x2y2+x2y2x3+y2\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 xx and yy .



qxpyp(x,y)=(6xy2+2xy6x2)2xyxy2x2qxpyp(x,y)=6x(y2x)x(y2x)=6\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 yy variable.

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



u(y)=e6dy=e6yu(y)=e^{\int 6dy}=e^{6y}

Back to our initial equation:



u(y)(p(x,y)dx+q(x,y)dy)=0e6y(xy2x2)dx+e6y(3x2y2+x2y2x3+y2)dy=0u(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)f(x,y)  that satisfied:



{fx=e6y(xy2x2)fy=e6y(3x2y2+x2y2x3+y2)fx=e6y(xy2x2)f(x,y)=(e6y(xy2x2))dx+g(y)f(x,y)=e6y(x2y22x33)+g(y)fy=y(e6y(x2y22x33)+g(y))==6e6y(x2y22x33)+e6y(x2y)+dgdy==e6y(3x2y22x3+x2y)+dgdy\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,


e6y(3x2y22x3+x2y)+dgdy=e6y(3x2y2+x2y2x3+y2)dgdy=e6yy2g(y)=(e6yy2)dy(integration by parts):g(y)=(e6yy2)dy=y2e6y62e6y6dy==y2e6y613(ye6y6e6y6)==y2e6y6ye6y18e6y108e6y(18y26y+1)108e^{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,



f(x,y)=f(x,y)=e6y(x2y22x33)+e6y(18y26y+1)108\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:



f(x,y)xdx+f(x,y)ydy=df(x,y)=0f(x,y)=Ce6y(x2y22x33)+e6y(18y26y+1)108=C\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



e6y(x2y22x33+18y26y+1108)=Ce^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}+\frac{18y^2-6y+1}{108}\right)=C


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