Question

Find the value of the constant r such that y=x^r satisfies the equation:
y'' - 3x^(-1)y' + 4x^(-2)y = 0

For this value of r verify that y = r^(x)log x is also a solution, what is the general solution?

Answer 1

\[y=x^{r} => y'=rx^{r-1}=>y''=r(r-1)x^{r-2}\]

Answer 2

plug in

Answer 3

\[r(r-1)x^{r-2}-3x^{-1}rx^{r-1}+4x^{-2}x^r\]
\[=r(r-1)x^{r-2}-3rx^{r-2}+4x^{r-2}\]
\[=x^{r-2}(r(r-1)-3r+4)\]
=0
so we have
\[r(r-1)-3r+4=0\]

Answer 4

ok I've solved for r and I get 2

Answer 5

\[r^2-r-3r+4=0\]
\[r^2-4r+4=0=> (r-2)^2=0=> r=2\]
very good!

Answer 6

Then I plug in \[2^{r}\ln x\]?

Answer 7

\[2^{x}\ln x\]

Answer 8

\[y=r^xlog(x)=r^x \frac{\ln(x)}{\ln(10)}\]
think:\[y=r^x => \ln(y)=\ln(r^x)=>\ln(y)=xln(r)=>y'=yln(r)=>y'=\ln(r) r^x\]
--------------
\[y'=\ln(r) r^x \frac{\ln(x)}{\ln(10)}+r^x \frac{1}{x} \frac{1}{\ln(10)}=\frac{1}{\ln(10)}r^{x}(\ln(r)\ln(x)+\frac{1}{x})\]
\[y''=\frac{1}{\ln(10)}(\ln(r)r^x(\ln(r)\ln(x)+\frac{1}{x})+\frac{1}{\ln(10)}r^x(\ln(r)\frac{1}{x}-\frac{1}{x^2})\]

Answer 9

now plug in

Answer 10

Flip, thats going to take a while to plug in!

Answer 11

What about the general solution, how would I go about finding it?

Answer 12

from the first part we have part of the solution is y=x^2
i don't feel like doing the 2nd part lol