Question

Find the particular solution of the differential equation
$(x-\sin y)dy+\left(\tan y\right)dx=0$, given that $y=0$, when $x=0$

Answer 1

We have,$(x-siny)dy+tanydx=0$

$\frac{dx}{dy}=-\left(\frac{x-siny}{tany}\right)$

$\frac{dx}{dy}+xcoty=cosy⟹\left(1\right)$

This is linear diff equation in the form,

$\frac{dx}{dy}+Rx=S$

Therefore the solution is given by

$x(I.F)=\int S(I.F)dy+C⟹\left(2\right)$

Where $I.F=e^{\int Rdy}$and $C$ is constant

Therefore $R=coty$ and $S=cosy$

$I.F=e^{\int Rdy}=e^{cotydy}=e^{logsiny}=siny$

Multiplying $\left(1\right)$ by $siny$

$\frac{dx}{dy}siny+xcotysiny=cosysiny$

$\frac{dx}{dy}siny+xcosy=cosysiny$

Integrating both sides,

$xsiny=\int cosysinydy+C$

$xsiny=\int \frac{sin2y}{2}dy+C$

$xsiny=\frac{1}{2}\int sin2ydy+C$

$xsiny=-\frac{cos2y}{4}+C⟹\left(3\right)$

When$x=0$and$y=0$

$0=-\frac{1}{4}+C$

$C=\frac{1}{4}$

Therefore $xsiny=\frac{{sin}^{2}y}{2}$

$2x=siny$

$y={sin}^{-1}2x$

We know that $-1\le 2x\le 1=\frac{-1}{2}\le x\le \frac{1}{2}$

$xE[\frac{-1}{2},\frac{1}{2}]$

Hence $y={sin}^{-1}\left(2x\right),xE[\frac{-1}{2},\frac{1}{2}]$is the solution.