Mathematics - Differential Equation Question with Solution | TestHub

Given, ${cosec}^{2}xdy+2dx=\left(1+y\cos 2x\right){cosec}^{2}xdx$

$\Rightarrow \frac{dy}{{\sin }^{2}x}+2dx=\left(\frac{1+y\cos 2x}{{\sin }^{2}x}\right)dx$

$\Rightarrow \frac{dy}{dx}+2{\sin }^{2}x=1+y\cos 2x$

$\Rightarrow \frac{dy}{dx}+2{\sin }^{2}x-1=y\cos 2x$

Using, $\cos 2x=1-2{\sin }^{2}x,$ we get

$\frac{dy}{dx}+\left(-\cos 2x\right)=y\cos 2x$

$\Rightarrow \frac{dy}{dx}+\left(-\cos 2x\right)y=\cos 2x$

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q,$ where $P\hspace{0.17em}\& Q$ are the functions of $x$ or constants.

Thus, $P=-\cos 2x \& Q=\cos 2x$

Now, we have integrating factor I.F.$=e^{\int Pdx}=e^{\int -cos2xdx}=e^{-\left(\frac{\sin 2x}{2}\right)}$

And, the solution of the given differential equation is

$y\left(I.F.\right)=\int Q\left(I.F.\right)dx+c$

 $\Rightarrow y·e^{-\left(\frac{\sin 2x}{2}\right)}=\int \left(\cos 2x\right)·e^{-\left(\frac{\sin 2x}{2}\right)}dx+c$

Put $\frac{\sin 2x}{2}=t, \Rightarrow \frac{2\cos 2x}{2}dx=dt$

$\Rightarrow y·e^{-\left(\frac{\sin 2x}{2}\right)}=\int e^{-t}dt+c$

$\Rightarrow y·e^{-\left(\frac{\sin 2x}{2}\right)}=-e^{-t}dt+c$

$\Rightarrow y·e^{-\left(\frac{\sin 2x}{2}\right)}=-e^{-\left(\frac{\sin 2x}{2}\right)}+c$

Given, $y\left(\frac{\pi }{4}\right)=0$

$\Rightarrow 0·e^{-\left(\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{2}\right)}=-e^{-\left(\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{2}\right)}+c$

$\Rightarrow c=-e^{-\left(\frac{1}{2}\right)}$

$\Rightarrow y·e^{-\left(\frac{\sin 2x}{2}\right)}=-e^{-\left(\frac{\sin 2x}{2}\right)}+e^{-\frac{1}{2}}$

Now, at $x=0$

$\Rightarrow y\left(0\right)·e^{-\left(\frac{\sin 0}{2}\right)}=-e^{-\left(\frac{\sin 0}{2}\right)}+e^{-\frac{1}{2}}$

$\Rightarrow y\left(0\right)·e^0=-e^0+e^{-\frac{1}{2}}$

$\Rightarrow y\left(0\right)=-1+e^{-\frac{1}{2}}$

$\Rightarrow y\left(0\right)+1=e^{-\frac{1}{2}}$

$\Rightarrow {\left(y\left(0\right)+1\right)}^2=e^{-1}.$