Question

show that the differential equation [ x sin² (y/x) - y] dx + xdy = 0 is homogenous. find the particular solution of this Differential equation, given that y = π/4 , when x = 1.​

Answer 1

Question:-

show that the differential equation [ x sin² (y/x) - y] dx + xdy = 0 is homogenous. find the particular solution of this Differential equation, given that y = π/4 , when x = 1.

Answer:-

Given differential equation is :

$\sf \left[ x\:\;sin^2\left(\dfrac{y}{x}\right)-y\right] dx + xdy = 0$

$\sf \implies \dfrac{dy}{dx} = \dfrac{y-x\:sin^2\left(\dfrac{y}{x}\right)}{x}\:\:\;\:\;\;\;\dots (i)$

let $\sf F(x,y) = \dfrac{y-x\:\:sin^{2}\left(\dfrac{y}{x}\right)}{x}$

on replacing x by λx and y by λy both sides, we get:

$\sf F(\lambda x , \lambda y) = \dfrac{\lambda\left[y-x\:\:sin^2\left(\dfrac{y}{x}\right)\right]}{\lambda x}$

$\sf = \lambda^{o} [ F(x,y)]$

thus, given differential equation is a homogeneous differential equation.

on putting y = vx

$\sf \implies \dfrac{dy}{dx} = v + x \: \dfrac{dv}{dx}$

on putting this value in eq. (1) we get

$\sf v + x \: \dfrac{dv}{dx} = \dfrac{vx-x\:\;sin^2\left(\dfrac{vx}{x}\right)}{x}$

$\sf \implies v + x \:\dfrac{dv}{dx} = v-sin^2v$

$\sf \implies x \dfrac{dv}{dx} = -sin^2v$

$\sf \implies cosec^2\:vdv = -\dfrac{dx}{x}$

integrating both sides, we get

$\displaystyle\sf\int\:cosec^2\:vdv\:+\:\int\dfrac{dx}{x}=0$

$\sf\implies -cot\:v + log|x| = C$

$\sf\implies -cot\left(\dfrac{y}{x}\right) + log |x| = C \:\:\left[\bf \because v = \dfrac{y}{x}\right]\:\;\;\;\dots(ii)$

$\rm$

Also, given that $\sf y = \dfrac{\pi}{4}\:\;\;when\;\;\; x = 1$

on putting these values in eq.(ii) we get

$\sf - cot \left(\dfrac{\pi}{4}\right) + log|1| = C$

$\sf \implies C = -1\;\;\;\;\bf\left[ \because cot \:\dfrac{\pi}{4} = 1\right]$

on putting the values of C in eq. (ii) , we get

$\sf -cot\left(\dfrac{y}{x}\right) + log|1| = -1$

$\boxed{\sf \implies 1 + log|1| - cot \left(\dfrac{y}{x}\right) = 0}$

which gives is the required particular solution of given differential equation.