Mathematics - Differential Equation Question with Solution | TestHub

We have,

$A=\left[\begin{array}{ccc}y& \sin x& 1\\ 0& -1& 1\\ 2& 0& \frac{1}{x}\end{array}\right]$

$\Rightarrow \left|A\right|=-\frac{y}{x}+2\sin x+2$

Now,

$\frac{dy}{dx}=\left|A\right|$

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

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

$\text{I.F.}=e^{\int \frac{1}{x}dx}=x$

Solution is

$y\times \left(\mathrm{I}.\mathrm{F}.\right)=\int \left(\mathrm{I}.\mathrm{F}.\right)\times \left(2\sin x+2\right)dx$

$\Rightarrow yx=\int x(2\sin x+2)dx$

$\Rightarrow yx=2\int x\left(\sin x\right)dx+2\int xdx$

$\Rightarrow xy=x^2-2x\cos x+2\sin x+c \dots \left(\mathrm{i}\right)$

Now $x=\pi , y=\pi +2$, hence

$\Rightarrow \mathrm{\pi }\left(\mathrm{\pi }+2\right)={\mathrm{\pi }}^2-2\mathrm{\pi }\left(-1\right)+0+\mathrm{c}$

$\Rightarrow c=0$

Now $\left(i\right)$ becomes

$xy=x^2-2x\cos x+2\sin x$

Put $x=\frac{\mathrm{\pi }}{2}$, then

$\left(\frac{\pi }{2}\right)y={\left(\frac{\pi }{2}\right)}^2-2·\frac{\pi }{2}\cos \left(\frac{\pi }{2}\right)+2\sin \left(\frac{\pi }{2}\right)$

$\Rightarrow \left(\frac{\pi }{2}\right)y=\frac{{\pi }^2}{4}+2$

$\Rightarrow y=\frac{\pi }{2}+\frac{4}{\mathrm{\pi }}$