Answer to Question #105574 in Differential Equations for Gayatri Yadav

Question

Solve the following differential equations:
i) y sin2x dx = (1+ y²+ cos²x) dy
ii) (xy² - x²)dx + (3x²y² + x²y - 2x³ + y²)dy = 0

Answer 1

i)

$ysin2xdx-(1+y^2+cos^2x)dy=0\\ P(x,y)=ysin2x, \\ Q(x,y)=-(1+y^2+cos^2x)\\ \frac{\partial P}{\partial y}=sin2x\\ \frac{\partial Q}{\partial x}=-2cosx\cdot (-sinx)=sin2x\\ \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$

equation in total differentials

$\exist u(x,y): \frac{\partial u}{\partial x}=P(x,y)\\ \frac{\partial u}{\partial y}=Q(x,y)\\$

In our case

$\frac{\partial u}{\partial x}=ysin2x\\ \frac{\partial u}{\partial y}=-(1+y^2+cos^2x)\\ u(x,y)=\int ysin2xdx=-\frac{y}{2}cos2x+\phi(y)$

Find

$\frac{\partial u}{\partial y}=-\frac{1}{2}cos2x+\phi'(y)=\\ =-\frac{1}{2}(2cos^2x-1)+\phi'(y)=\frac{1}{2}-cos^2x+\phi'(y)$

Then

$\frac{1}{2}-cos^2x+\phi'(y)=-1-y^2-cos^2x\\ \phi'(y)=-\frac{3}{2}-y^2\\ \phi(y)=-\frac{3}{2}y-\frac{y^3}{3}\\ u(x,y)=-\frac{y}{2}cos2x-\frac{3}{2}y-\frac{y^3}{3}$

Answer: solution of equation is

$-\frac{y}{2}cos2x-\frac{3}{2}y-\frac{y^3}{3}=C$

ii)

$(xy^2-x^2)dx+(3x^2y^2+x^2y-2x^3+y^2)dy=0\\ P(x,y)=xy^2-x^2\\ Q(x,y)=3x^2y^2+x^2y-2x^3+y^2\\ \frac{\partial P}{\partial y}=2xy\\ \frac{\partial Q}{\partial x}=6xy^2+2xy-6x^2\\ \frac{\partial P}{\partial y}\neq\frac{\partial Q}{\partial x}$

Find function $\mu(y)$

$\mu(y)=e^{\int F(y)dy}$

where

$F(y)=\frac{\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}}{-P}=\\ =\frac{2xy-6xy^2-2xy+6x^2}{x^2-xy^2}=\frac{6(x^2-xy^2)}{x^2-xy^2}=6\\ \mu(y)=e^{\int6dy}=e^{6y}$

we multiply the equation by $e^{6y}$

$(xy^2-x^2)e^{6y}dx+\\ +(3x^2y^2+x^2y-2x^3+y^2)e^{6y}dy=0\\ P_1(x,y)=(xy^2-x^2)e^{6y}\\ Q_1(x,y)=(3x^2y^2+x^2y-2x^3+y^2)e^{6y}\\ \frac{\partial P_1}{\partial y}=2xye^{6y}+6e^{6y}(xy^2-x^2)=\\ =e^{6y}(2xy+6xy^2-6x^2)\\ \frac{\partial Q_1}{\partial x}=(6xy^2+2xy-6x^2)e^{6y}\\ \frac{\partial P_1}{\partial y}=\frac{\partial Q_1}{\partial x}$

equation in total differentials

$\exist u(x,y): \frac{\partial u}{\partial x}=P_1(x,y)\\ \frac{\partial u}{\partial y}=Q_1(x,y)\\$

in our case

$\frac{\partial u}{\partial x}=(xy^2-x^2)e^{6y}\\ \frac{\partial u}{\partial y}=(3x^2y^2+x^2y-2x^3+y^2)e^{6y}\\ u=\int((xy^2-x^2)e^{6y})dx=(\frac{x^2y^2}{2}-\frac{x^3}{3})e^{6y}+\phi(y)$

Find

$\frac{\partial u}{\partial y}=x^2ye^{6y}+(\frac{x^2y^2}{2}-\frac{x^3}{3})6e^{6y}+\phi'(y)$

then

$x^2ye^{6y}+(\frac{x^2y^2}{2}-\frac{x^3}{3})6e^{6y}+\phi'(y)=\\ =(3x^2y^2+x^2y-2x^3+y^2)e^{6y}\\ \phi'(y)=y^2e^{6y}\\ \phi(y)=\int y^2e^{6y}dy=\frac{1}{6}\int y^2de^{6y}=\\ =\frac{1}{6}(y^2e^{6y}-\int 2ye^{6y}dy)=\frac{1}{6}(y^2e^{6y}-\frac{1}{3}\int yde^{6y})=\\ =\frac{1}{6}(y^2e^{6y}-\frac{1}{3}(ye^{6y}-\int e^{6y}dy))=\\ =\frac{1}{6}y^2e^{6y}-\frac{1}{18}ye^{6y}+\frac{1}{108}e^{6y}$

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

Answer: solution of equation is

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