Help with this : dy/dx+y/x=y^2*sinx

dy/dx+y/x=y^2*sinx
y^-2*(dy/dx)+y^-1*x^-1=sinx
let y^-1=t
=>dt/dx=-y^-2dy/dx
Substitute this in main eqn. ,
=>-dt/dx+t/x=sinx
=>dt/dx-t/x=-sinx
This is a linear differential equation so
I.F.=e^∫-(1/x)dx
I.F.=(1/x)
thus,
t*(1/x)=∫(-sinx)*1/xdx + c
Use back t=1/y
=>1/(xy)+∫(sinx/x)dx=c


Hope this helps :)

This is a nonlinear first order ODE. Looks like you'll need to do a u-substitution to turn this into a linear ODE and then solve by integrating factor, SO:

y' + (1/x)y = (y^2)sinx

u-substitution:

u = y^(-1) <----because u = y^(1 - n) and n = 2 in this case (y^2)
y = u^(-1) <----substitute this into ODE
y^2 = u^(-2) <------substitute this into ODE

dy/dx = (-u^(-2))(du/dx) <----substitute this into ODE as well

(-u^(-2))(du/dx) + (1/x)(u^(-1)) = u^(-2)(sinx)
du/dx + [(1/x)(u^(-1)) ]/(-u^(-2)) = u^(-2)(sinx)/(-u^(-2)) <----dividing everything by -u^(-2)
du/dx - (1/x)u = -sinx

Now it's a linear first order ODE, so we can do integrating factor:

p(x) = -(1/x)
q(x) = -sinx

z = e^[ integral: { p(x)dx } ]
z = e^[ integral: { -(1/x)dx } ]
z = e^[ ln[ 1/x ] ]
z = 1/x

So:

d[z*u]/dx = z*q(x)
d[z*u] = z*q(x)dx
(1/x)u = integral: -{ (1/x)sinxdx }

integral: -{ (1/x)sinxdx } <----This integral can be approximated using a power series representation. At this point the ODE gets ridiculously complicated to solve but it can be done.