Answer to Question #268598 in Differential Equations for Jes

Question

y(2x-y+1)dx + x(3x-4y+3)dy=0

Answer 1

$y(2x-y+1)dx+x(3x-4y+3)dy=0$

$M(x,y)dx+N(x,y)dy=0$

$M=y(2x-y+1)=2xy-y^2+y$

$N=x(3x-4y+3)=3x^2-4xy+3x$

$\frac{dM}{dy}=2x-2y+1$

$\frac{dN}{dx}=6x-4y+3$

$\frac{dM}{dy}\not=\frac{dN}{dx}$ ,thus the equation is not exact

$\frac{1}{M}(\frac{dN}{dx}-\frac{dM}{dy})=\frac{6x-4y+3-2x+2y-1}{y(2x-y+1)}$

$=\frac{4x-2y+2}{y(2x-y+1)}$

$=\frac{2(2x-y+1)}{y(2x-y+1)}=\frac{2}{y}$ which is a function of y

I.F=$e^{\int\frac{2}{y}dy}=e^{2lny}=y^2$

Multiplying both sides of the equation by y² we get

$y^3(2x-y+1)dx+xy^2(3x-4y+3)dy=0$

$(2xy^3-y^4+y^3)dx+(3x^2y^2-4xy^3+3xy^2)dy=0$

$\int(2xy^3-y^4+y^3)dx=x^2y^3-xy^4+xy^3, taking\ y\ as\ a\ constant$

$\int(3x^2y^2-4xy^3+3xy^2)dy=x^2y^3-xy^4+xy^3, taking\ x\ as\ a\ constant$

Adding the two, the solution of the equation is $x^2y^3-xy^4+xy^3=C$

Where C is an arbitrary constant