trouble with change of variables

[NHgirl]

Hi! Hoping someone can help me.
I have the following problem:
x^2y"+xy'-9y=0 ; use a change of variables of the form x=e^t
y(1)=0 and y'(1)=1

Although I know x^2=e^2t I don't see how that makes this equation any simpler to solve. I used the Cauchy-Euler equation and made some headway:
?^2-(1-1)?+9=0
?1,2=((1-1)±??)/2
?=-36 so ?1,2 = ±3i
which gives me:
y=C1cos(3t)+C2sin(3t)
I run into a little trouble here because y(1)=0 does not give me an easy cancellation to solve for C1. With both initial conditions evaluated I THINK the answer may be something like this:
y=-1/4*cos(3t)+1/4sin(3t)

Would someone be able to tell me if I'm close. If so, how would I solve this problem using x=e^t instead?? Thanks for any advice

 

[Deleted member 4993]

Hi! Hoping someone can help me.
I have the following problem:
x^2y"+xy'-9y=0 ; use a change of variables of the form x=e^t
y(1)=0 and y'(1)=1

Although I know x^2=e^2t I don't see how that makes this equation any simpler to solve. I used the Cauchy-Euler equation and made some headway:
?^2-(1-1)?+9=0
?1,2=((1-1)±??)/2
?=-36 so ?1,2 = ±3i
which gives me:
y=C1cos(3t)+C2sin(3t)
I run into a little trouble here because y(1)=0 does not give me an easy cancellation to solve for C1. With both initial conditions evaluated I THINK the answer may be something like this:
y=-1/4*cos(3t)+1/4sin(3t)

Would someone be able to tell me if I'm close. If so, how would I solve this problem using x=e^t instead?? Thanks for any advice

dy/dx = dy/dt * dt/dx = y[sub:1fhx2rv2]t[/sub:1fhx2rv2] * e[sup:1fhx2rv2]-t[/sup:1fhx2rv2]

y" = d/dt[y[sub:1fhx2rv2]t[/sub:1fhx2rv2] * e[sup:1fhx2rv2]-t[/sup:1fhx2rv2]]*dt/dx = [y[sub:1fhx2rv2]tt[/sub:1fhx2rv2] * e[sup:1fhx2rv2]-t[/sup:1fhx2rv2] - y[sub:1fhx2rv2]t[/sub:1fhx2rv2] * e[sup:1fhx2rv2]-t[/sup:1fhx2rv2]] * e[sup:1fhx2rv2]-t[/sup:1fhx2rv2] = e[sup:1fhx2rv2]-2t[/sup:1fhx2rv2][y[sub:1fhx2rv2]tt[/sub:1fhx2rv2] - y[sub:1fhx2rv2]t[/sub:1fhx2rv2]]

x^2y"+xy'-9y=0

[y[sub:1fhx2rv2]tt[/sub:1fhx2rv2] - y[sub:1fhx2rv2]t[/sub:1fhx2rv2]] + y[sub:1fhx2rv2]t[/sub:1fhx2rv2] - 9y = 0

y[sub:1fhx2rv2]tt[/sub:1fhx2rv2] - 9y = 0 with initial conditions y(0)=0 and y'(0)=1

This is now a simple ODE. Solve it.....

 

[tremor]

is this better?

y = 1/6*e^(3x-3)-1/6*e^(3x+3)

 

[tremor]

sorry to jump in on the thread btw...i'm trying to solve a similar problem

my apologies

 

[Deleted member 4993]

is this better?

y = 1/6*e^(3x-3)-1/6*e^(3x+3)

Incorrect..

Your independant variable now is 't'.

Then you can substitute back 'x'

 

[NHgirl]

I'm afraid I don't understand the ytt notation. I'm a little lost :?: Could you explain the x substitution further?

 

[Deleted member 4993]

I'm afraid I don't understand the ytt notation. I'm a little lost :?: Could you explain the x substitution further?

\(\displaystyle \frac{dy}{dt} \ = \ y_t\)

\(\displaystyle \frac{d^2y}{dt^2} \ = \ y_{tt}\)

 

[NHgirl]

Ok I figured as much. So it would be incorrect to solve for:

y''-9y=0

because I have not substituted back x? How would I do that? Or does the answer just have to be in terms of t rather than x?

 

[Deleted member 4993]

Ok I figured as much. So it would be incorrect to solve for:

y''-9y=0 <<< Yes that would be incorrect for this problem since your original problem was x[sup:2oi4r9o8]2[/sup:2oi4r9o8]y"+xy'-9y=0 ...

because I have not substituted back x? How would I do that? Or does the answer just have to be in terms of t rather than x?

Your interim solution of the ODE will be interms of t - your new independant variable.

Then you'll replace "t" using the equation x = e[sup:2oi4r9o8]t[/sup:2oi4r9o8] - for the final solution.

 

[NHgirl]

ok sorry to be a pain and hopefully this will be the last post but...

if i get an interim solution of:

y=C1e^3t+C2e^-3t

do i substitute x back in as:

y=C1x^3+C2(1/x^3)?

i'm just trying to understand the whole concept. i very much appreciate your help!

 

[Deleted member 4993]

ok sorry to be a pain and hopefully this will be the last post but...

if i get an interim solution of:

y=C1e^3t+C2e^-3t

do i substitute x back in as:

y=C1x^3+C2(1/x^3) <<<< Correct?

i'm just trying to understand the whole concept. i very much appreciate your help!