Question

\[x^2y^{\prime\prime}+2xy^\prime-2y=x^3\]

Answer 1

how to start?

Answer 2

Euler equation
let t=ln x

Answer 3

true

Answer 4

it will gives
\[y_{t}''+y_{t}'-2y_{t}=e^{3t}\]

Answer 5

give***

Answer 6

for t
\[D^2+D-2=0\\ D=2 \ ,-1\]
so
\[y_{c}=c_{1}e^{2t}+c_{2}e^{-t}\]
choosing
\[y_{p}=Ae^{3t}\]
gives
\[A=\frac{1}{10}\]
then we have
\[y_{t}=c_{1}e^{2t}+c_{2}e^{-t}+ \frac{1}{10} e^{3t}\]
convert t to x
\[y_{x}=c_{1} x^2 +\frac{c_{2}}{x}+ \frac{1}{10} x^3\]

Answer 7

Brilliant.

Answer 8

\[x^2y^{\prime\prime}+2xy^\prime-2y=x^3\]
\[\text {let}\quad t=\ln x \]\[e^t=x\]
\[e^{2t}y^{\prime\prime}+2e^ty^\prime-2y=e^{3t}\]
\[y^{\prime\prime}+(2-1)y^\prime-2y=e^{3t}\]
\[y^{\prime\prime}+y^\prime-2y=e^{3t}\]
\[y_c^{\prime\prime}+y_c^\prime-2y=0\]\[m^2+m-2=0\]
\[(m-1)(m+2)=0\]\[m=2,-1\]\[y_c=Ae^{-t}+Be^{2t}\]
\[y_p^{\prime\prime}+y_p^\prime-2y_p=e^{3t}\]\[y_p=Ce^{3t};\qquad\qquad y^\prime_p=3Ce^3t;\qquad\qquad y^{\prime\prime}_p=9Ce^3t\]\[9Ce^{3t}+3Ce^{3t}-2Ce^{3t}=e^{3t}\]\[10Ce^{3t}=e^{3t}\]\[C=\frac 1{10}\]
\[y_p=\frac 1{10}e^{3t}\]

\[y(t)=Ae^{-t}+Be^{2t}+\frac 1{10}e^{3t}\]
\[y(x)=Ax^{-1}+Bx^{2}+\frac 1{10}x^{3}\]