Question

Can someone please check my work
\[x^2y^{\prime\prime}-3xy^\prime+3y=0;\qquad y(1)=5;\qquad y(1)^{\prime\prime}=12\]

Answer 1

\[x^2y^{\prime\prime}-3xy^\prime+3y=0;\qquad y(1)=5;\qquad y(1)^{\prime\prime}=12\]
\[\text{let } \ln x=t\]\[\qquad x=e^t\]
\[y^{\prime\prime}+(-3-1)y^\prime+3y=0\]
\[y^{\prime\prime}-4y^\prime+3y=0\]\[m^2-4m+3=0\]\[(m-1)(m-3)=0\]\[m=1,3\]
\[y(t)=Ae^t+Be^{3t}\]
\[y(x)=Ax+Bx^3\]\[y^\prime(x)=x+3Bx^2\]\[y^{\prime\prime}(x)=6Bx\]\[y^{\prime\prime}(1)=6B=12;\qquad B=2\]\[y(x)=Ax+2x^3\]\[y(1)=A+2=5;\qquad A=3\]
\[y(x)=3x+2x^3\]

Answer 2

Yes.

Answer 3

YES

Answer 4

But you may want to explain your leap from
\[\text{let }\ln x=t\ \ x=e^t\]to\[y''+(−3−1)y'+3y=0\]

Answer 5

can you help with that im not really sure of the details

Answer 6

\[x=e^t\]\[\frac{\text dx}{\text dt}=e^t\]\[\text dx=e^t\text dt\]