Question

ydx + (x y +x-3y) dy=0
Solve The differential eqn by
Integrating factor​

Answer 1

We are given to solve the differential equation,

$\longrightarrow y\ dx+(xy+x-3y)\ dy=0$

Dividing by $dy,$

$\longrightarrow y\ \dfrac{dx}{dy}+xy+x-3y=0$

$\longrightarrow y\ \dfrac{dx}{dy}+x+(x-3)y=0$

Dividing by $y,$

$\longrightarrow\dfrac{dx}{dy}+\dfrac{x}{y}+x-3=0$

$\longrightarrow\dfrac{dx}{dy}+x\left(1+\dfrac{1}{y}\right)=3$

Now we have a first order linear differential equation of the form,

$\longrightarrow\dfrac{dx}{dy}+x\cdot P(y)=Q(y)$

where,

• $P(y)=1+\dfrac{1}{y}$

• $Q(y)=3$

So,

$\displaystyle\longrightarrow\int P(y)\ dy=\int\left(1+\dfrac{1}{y}\right)\ dy$

$\displaystyle\longrightarrow\int P(y)\ dy=y+\log|y|$

Then integrating factor,

$\displaystyle\longrightarrow IF=e^{\int P(y)\ dy}$

$\displaystyle\longrightarrow IF=e^{y+\log|y|}$

$\displaystyle\longrightarrow IF=ye^y$

The solution of the differential equation is given by,

$\displaystyle\longrightarrow x\cdot IF=\int Q(y)\cdot IF\ dy$

$\displaystyle\longrightarrow xye^y=\int3ye^y\ dy$

$\displaystyle\longrightarrow xye^y=3\left[ye^y-\int e^y\ dy\right]$

$\displaystyle\longrightarrow xye^y=3\left(ye^y-e^y\right)+C$

$\displaystyle\longrightarrow xye^y=3e^y\left(y-1\right)+C$

$\displaystyle\longrightarrow\underline{\underline{xy=3\left(y-1\right)+Ce^{-y}}}$