Question

solve x²y'' =2xy'+(y')²​

Answer 1

Step-by-step explanation:

Can anyone give me an advice that helps me to solve this kind of DE:

x2⋅y′′+2x⋅y′−2y=0

knowing that

y1=Ax+Bx2

is a solution.

I've tried to solve it by reduction of order (described here) by assuming that

y2=y1⋅v

is a solution where v(x) is a function.

I've calculated y′′2 and y′2 then I've tried using them in the original DE and I got:

v′′(Ax3+B)+v′(4Ax2−2Bx)=0

Next I assumed that there exists a function w such that w=v′ and w′=v′′

so the equation above reduced to the first order DE:

w′(Ax3+B)+w(4Ax2−2Bx)=0

My intuition tells me that I should guess a function u(x) such that the DE above writes like

w′⋅u+w⋅u′=0⟺(w⋅u)′=0⟹w⋅u=C

where u(x) could be something like:

u=43⋅Ax3−2B⋅ln(x)

but it doesn't work because:

u≠(Ax3+B)

Answer 2

ANSWER:

ok.......................................