Question

1.1) y^-2 dy/dx + y^-1 = 2x

1.2) (y^2e^xy^2 +4x^3)dx + (2xye^xy^2 - 3y^2)dy = 0

Answer 1

$1.1) \ \frac{1}{y^2} \frac{dy}{dx}+\frac{1}{y}=2x$

$\ -\frac{1}{y^2} \frac{dy}{dx}-\frac{1}{y}=-2x$

$v(x)=\frac{1}{y(x)}, \ \frac{dv}{dx}=-\frac{1}{y^2(x)}\frac{dy}{dx}$

Substitute $v(x)$ :

$\frac{dv }{dx}-v=-2x$

Multiply by $e^{-x}$ both sides of equation:

$e^{-x}\frac{dv}{dx} -e^{-x}v=-2xe^{-x}$

$e^{-x}\frac{dv}{dx} +v\frac{d}{dx}(e^{-x})=-2xe^{-x}$

$\frac{d}{dx}(ve^{-x})=-2xe^{-x}$

$ve^{-x}=\int (-2xe^{-x})dx=2xe^{-x}+2e^{-x}+C$

Return to $y(x):$

$\frac{1}{y}e^{-x}= 2xe^{-x}+2e^{-x}+C$

$y=\frac{e^{-x}}{2xe^{-x}+2e^{-x}+C}=\frac{1}{2x+2+Ce^{x}}$

Answer: $y(x)=\frac{1}{Ce^x+2(x+1)}$

$1.2) (y^2e^{xy^2}+4x^3)dx+(2xye^{xy^2}-3y^2)dy=0$

$P(x,y)=y^2e^{xy^2}+4x^3, \ \ Q(x,y)=2xye^{xy^2}-3y^2$

$\frac{\partial P(x,y)}{\partial y} =\frac{\partial Q(x,y)}{\partial x}=2ye^{xy^2}+2xy^3e^{xy^2}$

So, we will find function $f(x,y):$

$\frac{\partial f(x,y)}{\partial x}=y^2e^{xy^2}+4x^3, \ \ \frac{\partial f(x,y) }{\partial y}=2xye^{xy^2}-3y^2$

$f(x,y)=\int (y^2e^{xy^2}+4x^3)dx= e^{xy^2}+x^4+\phi (y)$

$f(x,y)=\int (2xye^{xy^2}-3y^2)dy =e^{xy^2} -y^3+\psi (x)$

$f(x,y)=e^{xy^2}+x^4-y^3$

$\frac{\partial f(x,y)}{\partial x} dx+\frac{\partial f(x,y)}{\partial y} dy=0 \ \Rightarrow f(x,y)\equiv C$

Answer: $e^{xy^2(x)}+x^4-y^3(x)=C$