Answer to Question #150871 in Differential Equations for RIZWAN KHAN

Question

Solve the given homogeneous differential equation
X2dy+y(x+y)dx=0

Answer 1

$x^2dy+y(x+y)dx=0$
1) Expand:x² dy+yx dx +y² dx = 0or$y'+ \dfrac{y}{x}+ \dfrac{y^2}{x^2}=0$ .....(*)   first-order nonlinear ODE.

2) We need to transform it into linear ODEObserve the exponent of y, it's n=2. So we'll use substitution $v=y^{1-n}=y^{1-2}=y^{-1}$
3) So we have$v= \dfrac{1}{y}$ $v'= (\dfrac{1}{y}) '$ $v'=-( \dfrac{1}{y^2} )y'=-v^2y'$ In (*) substitute $y'=-( \dfrac{1}{v^2} )v'$ and $y= \dfrac{1}{v}$ $\dfrac{-v}{v^2} + \dfrac{1}{xv}+ \dfrac{1}{x^2v^2}=0$ .... (**)
this is a first-order linear ODE.4) New substitutionv = x uwhere u=u(x) and that's all we need to know about "u" for nowv' = xu' + uback to (**)$- \dfrac{(xu'+u)}{x^2u^2} + \dfrac{1}{x^2u} + \dfrac{1}{(x^4u^2)} =0$ or expanded$- \dfrac{u'}{xu^2} - \dfrac{1}{x^2u} +\dfrac{1}{(x^2u)}+ \dfrac{1}{(x^4u^2)} =0$ 5) Solve for u' 
two middle terms are canceled:$-\dfrac{u'}{(xu^2)}+\dfrac{1}{(x^4u^2)}=0$
multiply by xu²$-u'+ \dfrac{1}{x^3} =0$ $u'= \dfrac{1}{x^3}$
6) Integrate with respect to x$u=\int( \dfrac{1}{x^3} )dx$ $u=- \dfrac{1}{2x^2}+C$

7) Substitute back $u= \dfrac{v}{x}$ $\dfrac{v}{x} =- \dfrac{1}{2x^2}+C$
8) Multiply by x$v=- \dfrac{1}{2x}+Cx$ 9) Substitute back $v= \dfrac{1}{y}$ $\dfrac{1}{y} =- \dfrac{1}{(2x)}+Cx$ $\dfrac{1}{y} =- \dfrac{1}{(2x)}+ \dfrac{2Cx^2}{2x}= \dfrac{(2Cx^2-1)}{2x}$

$y= \dfrac{2x}{2Cx^2-1}$

Since C is constant, for simplicity of solution we can introduce a new constantC₁ = 2C so finally$y= \dfrac{2x}{(C_1x^2-1)}$