Reduction of order
DE: x^2y''-7xy'+16y=0
y1=x^4

I set y2 equal to v(x)x^4, plugged in and got it to just v' and v'' terms, substituted w and i have wx^6+w'x^5=0
I'm guessing this is a really easy DE to solve because its 1st order, but i know its not linear because its equal to 0.

Question

Reduction of order
DE: x^2y''-7xy'+16y=0
y1=x^4

I set y2 equal to v(x)x^4, plugged in and got it to just v' and v'' terms, substituted w and i have wx^6+w'x^5=0
I'm guessing this is a really easy DE to solve because its 1st order, but i know its not linear because its equal to 0.

anonymous · Answer

Use the substitution $x=e^t$.

anonymous · Answer

This way you'll get a second order DE with constant coefficients.

jamesj · Answer

Does the question ask you to reduce order?  It's a slightly painful way to solve this problem.

jamesj · Answer

Which is to say, if you start with your original equation,
$$ x^2y''-7xy'+16y=0 $$
If you use a trial solution/Ansatz of $ y = x^r $, you'll find the solution.

anonymous · Answer

No it didn't say linear thats just the one I first wanted to use when i saw it. I know i don't want second order, its a reduction of order problem. I'm guessing i need to get them into dw and dx, is w' the same as dw/dx? Or can I decide that's what it represents?

anonymous · Answer

@James: It worked with me. I've got $y''(t)-8y'(t)+16y(t)=0$, where u is a function of t=ln(x). Am I right here?

anonymous · Answer

y*

anonymous · Answer

y2 is supposed to be 1

anonymous · Answer

oops looking at the answer to the next problem, no idea what y2 is

anonymous · Answer

If this is right, then the auxiliary equation is $m^2-8m+16=0 \implies (m-4)^2=0 \implies m=4$. So we have a repeated root at m=4. So $y(t)=c_1e^{4t}+c_2xe^{4t}$. Npw, plug t=ln x.

anonymous · Answer

Then $y(x)=c_1e^{4 \ln{x}}+c_2\ln(x)e^{4\ln{x}}=c_1x^4+c_2\ln(x)x^4.$

anonymous · Answer

So yeah $x^4$ is your first solution, and $\ln(x)x^4$ is your second solution.

anonymous · Answer

I made a typo on above, it should be $y(t)=c_1e^{4t}+c_2te^{4t}.$ The final answer is right.

anonymous · Answer

typo above*

anonymous · Answer

thats the best answer i've heard so far...not the way we were taught to do it but i ended up with -x^4ln(x) by seperating the variables once i got it into W's. if someone could tell me if the way i did it is ok i'd really appreciate it.

(dw/dx)*x^6 = -w*x^5
dw/w = -dx/x
ln(w) = -ln(x) or ln(1/x)
w = 1/x
v = ln(x)

realized while typing it i changed the log to 1/x and kept the negative. thanks guys.

anonymous · Answer

I think what you did is correct.