Answer to Question #183544 in Differential Equations for Genevieve Lugo

Question

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

Answer 1

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

Comparing with Mdx+Ndy=0;

where M=y(x+y) , N=x+2y-1.

$\therefore\;\frac{\partial M}{\partial y} =x+2y\; and\;\frac{\partial N}{\partial x}=1\newline$

It is a non exact differential equation beacuse $\frac{\partial M}{\partial y} \neq\frac{\partial N}{\partial x}$

Integrating factor:

$\exp( \int \frac{\frac{\partial M}{\partial x} -\frac{\partial N}{\partial y} }{N}dx)=\exp\int1dx\newline I.F = \exp(x)$

Multiplying I.F throughout the differential equation.

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

Now this differential equation is exact.

General solution of a exact differential equation is given by

$\int Mdx(considering\;y\;constant) + \int Ndy(term \;which\; do\; not\; contain \;x )=constant$

$\int(\exp(x)xy+\exp(x)y^2)+\int(0)dy=constant$

$y\int \exp(x)x+y^2\int\exp(x)=constant(k)\newline y\cdot\exp(x)\cdot(x-1)+y^2\cdot\exp(x)=k$

Thus general solution of above differential equation is

$y^2\cdot\exp(x)+y\cdot\exp(x)\cdot(x-1)=k$