Question

Find the particular solution of the differential equation $\frac{dx}{dy}+xcoty=2y+y^{2}coty,(y\ne 0)$, given that $x=0$ when $y=\frac{\pi }{2}$.

Answer 1

We have $\frac{dx}{dy}+xcoty=2y+y^{2}coty$, $(y\ne 0)$
It is clear that this ia linear differential equation of the form $\frac{dx}{dy}+P\left(y\right)x=Q\left(y\right)$
Here, $P\left(y\right)=coty,Q\left(y\right)=2y+y^{2}coty$
Integrating factor = $e^{\int cotydy}=e^{logsiny}=siny$
So, solution is given by, $xsiny=\int siny(2y+y^{2}coty)dy$
$xsiny=\int 2ysinydy+\int y^{2}cosydy$
$xsiny=\int 2ysinydy+y^2\int cosydy-\int (\frac{d\left(y^2\right)}{dy}\int cosydy)dy$
$xsiny=\int 2ysinydy+y^{2}siny-\int 2ysinydy$
$xsiny=y^{2}siny+C$
Given that x = 0, when $y=\frac{\pi }{2}$, we get $0sin\frac{\pi }{2}=(\frac{\pi }{2}{)}^{2}sin\frac{\pi }{2}+c$
Therefore, $c=-\frac{{\pi }^2}{4}$
So, the required solution is: $xsiny=y^{2}siny-\frac{{\pi }^2}{4}$