3.3.14.1 Example

\begin{equation} y^{\prime }=5e^{x^{2}+20y}+\sin x \tag {1}\end{equation}

Here \(a\left ( x\right ) =x^{2},b=20,f\left ( x\right ) =\sin x,g\left ( x\right ) =5\). Hence let

\begin{align*} u & =e^{-by}\\ & =e^{-20y}\end{align*}

Therefore

\begin{align*} \frac {du}{dx} & =-20y^{\prime }e^{-20y}\\ & =-20y^{\prime }u \end{align*}

Or

\begin{equation} y^{\prime }=-\frac {u^{\prime }}{20u} \tag {2}\end{equation}

Comparing (1,2) gives

\begin{align*} -\frac {u^{\prime }}{20u} & =5e^{x^{2}+20y}+\sin x\\ & =5e^{20y}e^{x^{2}}+\sin x\\ & =5\frac {1}{u}e^{x^{2}}+\sin x \end{align*}

Or

\begin{align} -u^{\prime } & =100e^{x^{2}}+20u\sin x\nonumber \\ u^{\prime } & =-100e^{x^{2}}-20u\sin x\nonumber \\ u^{\prime }+20u\sin x & =-100e^{x^{2}} \tag {3}\end{align}

This is linear first order ode. The integrating factor is

\begin{align*} I & =e^{\int 20\sin xdx}\\ & =e^{-20\cos x}\end{align*}

(3) becomes

\begin{align*} \frac {d}{dx}\left ( uI\right ) & =-I100e^{x^{2}}\\ ue^{-20\cos x} & =-100\int e^{x^{2}}e^{-20\cos x}dx+c\\ u & =-100e^{20\cos x}\int e^{x^{2}-20\cos x}dx+ce^{20\cos x}\\ & =e^{20\cos x}\left ( -100\int e^{x^{2}-20\cos x}dx+c\right ) \end{align*}

But \(u=e^{-20y}\) therefore

\begin{align*} e^{-20y} & =e^{20\cos x}\left ( -100\int e^{x^{2}-20\cos x}dx+c\right ) \\ -20y & =\ln \left ( e^{20\cos x}\left ( -100\int e^{x^{2}-20\cos x}dx+c\right ) \right ) \\ y & =-\frac {1}{20}\ln \left ( e^{20\cos x}\left ( -100\int e^{x^{2}-20\cos x}dx+c\right ) \right ) \end{align*}