Question

y(x + y)dx + (x + 2y − 1)dy = 0

Answer 1

Solution.

$y(x+y)dx+(x+2y−1)dy=0,\newline (yx+y^2)dx+(x+2y−1)dy=0.\newline \text{Let be } M(x,y)=yx+y^2, N(x,y)=x+2y-1.$

Find integrating multiplier $\mu(x,y):$

$\frac{M_y-N_x}{N}=\frac{x +2y-1}{x+2y-1}=1.$

$\mu(x,y)=e^{\int 1dx}=e^x.$

We will have

$(yx+y^2)e^xdx+(x+2y−1)e^xdy=0.\newline$

This equation is exact. Then $F(x,y)=\int (xy+y^2)e^xdx=y^2e^x+(xe^x-e^x)y+g(y),$

from here $F_y=2ye^x+xe^x-e^x+g'(x)=N(x,y)=(x+2y-1)e^x,$

then $g'(x)=0, g(x)=C,$ where $C$ is some constant.

So, $e^xy^2+(xe^x-e^x)y+C=0$ is the solution of equation.

Answer. $e^xy^2+(xe^x-e^x)y+C=0.$