Question

the integrating factor for the differential equation (dy/dx)+PY=Q​

Answer 1

Given a differential equation,

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

Multiply both sides by $I(x)$ which is called integrating factor, it's a function in $x.$

$\longrightarrow I(x)\cdot\dfrac{dy}{dx}+I(x)\cdot P(x)\cdot y=I(x)\cdot Q(x)\quad\quad\dots(1)$

Assume,

$\longrightarrow\dfrac{d}{dx}\left[I(x)\right]=I(x)\cdot P(x)\quad\quad\dots(2)$

To satisfy this condition, $I(x)$ should be in the form of $e^{f(x)},$ where $f(x)$ is a function in $x.$

So let $I(x)=e^{f(x)}.$ Then,

$\longrightarrow\dfrac{d}{dx}\left[e^{f(x)}\right]=e^{f(x)}\cdot P(x)$

$\longrightarrow e^{f(x)}\cdot f'(x)=e^{f(x)}\cdot P(x)$

$\longrightarrow f'(x)=P(x)$

$\displaystyle\longrightarrow f(x)=\int P(x)\ dx$

Hence the integrating factor,

$\longrightarrow\underline{\underline{I(x)=e^{\int P(x)\ dx}}}$

Well I'm keeping on solving the differential equation.

Putting (2) in (1),

$\longrightarrow I(x)\cdot\dfrac{dy}{dx}+\dfrac{d}{dx}\left[I(x)\right]\cdot y=I(x)\cdot Q(x)$

By product rule,

$\longrightarrow\dfrac{d}{dx}\left[I(x)\cdot y\right]=I(x)\cdot Q(x)$

$\displaystyle\longrightarrow I(x)\cdot y=\int I(x)\cdot Q(x)\ dx\quad\!(+C)$

That is,

$\displaystyle\longrightarrow y\cdot e^{\int P(x)\ dx}=\int Q(x)\cdot e^{\int P(x)\ dx}\ dx\quad\!(+C)$

where $C$ is the integral constant.