# please sole the following problemprove that the differential eqn x^2dy/dx-xy=1+cos(x/y) , xnot=0 is homogeneous. find the particular solution of tis differential eqn given that when x=1 y=pi/2

Jitender Singh IIT Delhi
10 years ago
Ans:
$x^{2}\frac{dy}{dx}-xy=1+cos(\frac{x}{y})$
Let x = uy
$1 = u\frac{dy}{dx}+y\frac{du}{dy}$
$\frac{dy}{dx}=\frac{1-y\frac{du}{dy}}{u}$
$(uy)^{2}\frac{1-y\frac{du}{dy}}{u}-uy^{2}=1+cosu$
$-uy^{3}\frac{du}{dy}=1+cosu$
$\int \frac{u}{1+cosu}du=\int \frac{-1}{y^{3}}dy = \frac{1}{2y^{2}}+c$
$\int \frac{u}{1+cosu}du=\int \frac{u}{2cos^{2}\frac{u}{2}}du=\frac{1}{2}\int u.sec^{2}\frac{u}{2}du$
Integration by parts
$=\frac{1}{2}[2utan\frac{u}{2}-\int 2tan\frac{u}{2}du] = utan\frac{u}{2}-2lnsec\frac{u}{2}$
$utan\frac{u}{2}-2lnsec\frac{u}{2} = \frac{1}{2y^{2}}+c$
$\frac{x}{y}tan\frac{x}{2y}-2lnsec\frac{x}{2y} = \frac{1}{2y^{2}}+c$
$x = \frac{\pi }{2}\rightarrow y = 1$
$\frac{\pi }{2}tan\frac{\pi }{4}-2lnsec\frac{\pi }{4} = \frac{1}{2}+c$
$c = \frac{\pi }{2}-2lnsec\sqrt{2} - \frac{1}{2}$
$c = \frac{\pi -1-ln4}{2}$
$\frac{x}{y}tan\frac{x}{2y}-2lnsec\frac{x}{2y} = \frac{1}{2y^{2}}+\frac{\pi -1-ln4}{2}$
Thanks & Regards
Jitender Singh
IIT Delhi