Answer :

Solution :

$(\sin <!--\; \; -->x)\frac{dy}{dx}+y\cos <!--\; \; -->x=1$

Dividing both sides by "sin x", the equation is reduced to

$\frac{dy}{dx}+y\cot <!--\; \; -->x=\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}$

The above equation is linear.

By comparing with standard linear equation i.e.

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

P(x) = cot x and Q(x) = cosec x.

We know that;

$I.F.=e^{\int <!--\; \int \; -->Pdx}$

$I.F.=e^{\int <!--\; \int \; -->\cot <!--\; \; -->xdx}=e^{\log <!--\; \; -->\sin <!--\; \; -->x}=\sin <!--\; \; -->x$

The solution is given by :

$y\left(I.F.\right)=\int <!--\; \int \; -->\left(Q\times I.F.\right)dx+c$

y × sin x = ∫(cosec x × sin x)dx + c

y × sin x = ∫dx + c

y sin x = x + c

Boundary condition: y(π/2) = π/2

$\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+c$

∴ c = 0.

y sin x = x

y(π/6) equals;

$y\sin <!--\; \; -->\left(\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}\right)=\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}$

$y\left(\frac{1}{2}\right)=\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}$

$\therefore <!--\; \therefore \; -->y=\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}$