Solve the homogeneous ode 2xy²dy-(x³+2y³dx=0?

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Answer 1 · 2018-05-10T14:16:26

#(2y^3)/(3x^3)=ln(cx)##ore^((2y^3)/(3x^3) =cx#

Here,

#2xy²dy-color(red)((x³+2y³)) dx=0#

#=>2xy²dy=(x³+2y³)dx#

#=>(dy)/(dx)=(x^3+2y^3)/(2xy^2)#

#=>(dy)/(dx)=(x^3(1+2(y/x)^3))/(x^3(2(y/x)^2)#

#=>(dy)/(dx)=(1+2(y/x)^3)/(2(y/x)^2)#

Let us subst.

#color(blue)(y/x=v=>y=vx=>(dy)/(dx)=v+x(dv)/(dx)#

So,

#v+x(dv)/(dx)=(1+2v^3)/(2v^2)#

#=>x(dv)/(dx)=(1+2v^3)/(2v^2)-v#

#=>x(dv)/(dx)=(1+2v^3-2v^3)/(2v^2)=1/(2v^2)#

#=>2v^2dv=1/xdx#

Integrating both sides :

#=>2intv^2dv=int1/xdx+c'#

#=>2*v^3/3=lnx+lnc...to[c'=lnc]#

#=>2/3v^3=ln(cx)#

Subst. back, #color(blue)(v=y/x#,we get

#2/3(y^3/x^3)=ln(cx)#

#=>(2y^3)/(3x^3)=ln(cx)#

#=>e^((2y^3)/(3x^3) =cx#