Thanks for pointing that out! I'll make sure to double check my work next time.

[Oxymoron]

Consider the differential equation

t^2x''-tx'+x=0

Given that x(t)=t is a solution of the ordinary differential equation, find another linearly independent solution.

\textsc{SOLUTION}

Since we know x(t) = t is a solution, the general solution of the ODE will be of the form x(t) = Kt + Kty(t), where y(t) is some function linearly independent to x(t) = t. This suggests that we utilize a solution of the form:

x(t) = ty(t)

Take x(t) = ty(t). Then

x'(t) = y(t) + ty'(t)
x''(t) = 2y'(t) + y''(t)

Substitute this back into the original ODE we get

t^2\left(2y'(t) + y''(t)\right) - t\left(y(t) + ty'(t)\right) + ty(t) = 0
t^2y''(t)+t^2y'(t) -ty(t) +ty(t) = 0
t^2y''(t)+t^2y'(t) = 0
t^2\left(y''(t)+y'(t)\right) = 0
y''(t) + y'(t) = 0

Integrating both sides with respect to t we obtain

y'(t) + y(t) = C

Rearranging

y'(t) = -y(t) + C

This differential equation is separable

\frac{dy(t)}{dt} = -y(t) + C

\frac{dy(t)}{(-y(t)+C)} = dt

\int \frac{dy(t)}{(-y(t)+C)} = \int dt

-\ln|y(t)+C| = t

-\ln|y(t)|\ln|C|=t

\ln\left|\frac{1}{y(t)}\right|C = t

\frac{1}{y(t)} = e^{tC'}

y(t) = e^{-tC'}

y(t) = C'e^{-t}

Therefore x(t) = ty(t) = C'te^{-t} is another linearly independent solution.

But when I use Maple (to check that this solution works), it says that y(t) = C'e^{-t} is not a solution. Instead is solves the ODE and finds that x(t) = t and x(t) = \ln|t| are the solutions. So I must have made a mistake somewhere??

 

[ehild]

Consider the differential equation

t^2x''-tx'+x=0

Given that x(t)=t is a solution of the ordinary differential equation, find another linearly independent solution.

This suggests that we utilize a solution of the form:

x(t) = ty(t)

Take x(t) = ty(t). Then

x'(t) = y(t) + ty'(t)
x''(t) = 2y'(t) + y''(t)

You missed a "t".

x''(t) = 2y'(t) + ty''(t)

ehild

 

[Oxymoron]

Thankyou SOOOOOOO Much. That is it. It all works now. :-)

 

[Oxymoron]

Well, kind of. I'm stuck again.

x(t) = ty(t)

Take x(t) = ty(t). Then

x'(t) = y(t) + ty'(t)
x''(t) = 2y'(t) + ty''(t)

Substitute this back into the original ODE we get

t^2\left(2y'(t) + ty''(t)\right) - t\left(y(t) + ty'(t)\right) + ty(t) = 0
t^3y''(t)+t^2y'(t) -ty(t) +ty(t) = 0
t^3y''(t)+t^2y'(t) = 0
t^2\left(ty''(t)+y'(t)\right) = 0
ty''(t) + y'(t) = 0

Rearranging

y''(t) +\frac{1}{t}y'(t) = 0

Integrating both sides with respect to t we obtain

\int y''(t) + \int\frac{1}{t}y'(t) = C

Cant solve this? Is there an easier way? Or am I missing something?

 

[dextercioby]

Make the substitution

y'(t)=u(t)

Daniel.

 

[Oxymoron]

\int u'(t)dt + \int \frac{1}{t}u(t)dt = 0

Integration by parts

u(t) + \left(\frac{1}{t}\left(\int u(t)\right)\right) - \left(\int u(t)\right)\left(-\frac{1}{t^2}\right) = 0

But \int u(t) = y(t)

y'(t) + \left[\frac{1}{t}y(t) - \left(y(t)\left(-\frac{1}{t^2}\right)\right)\right] = 0

y'(t) + \left[\frac{1}{t}y(t) + \frac{1}{t^2}y(t)\right] = 0

y'(t) + y(t)\left(\frac{1}{t} + \frac{1}{t^2}\right) = 0

y'(t) + y(t)\left(\frac{t+1}{t^2}\right) = 0

Am I going ok?

 

[ehild]

Integration by parts

u(t) + \left(\frac{1}{t}\left(\int u(t)\right)\right) - \left(\int u(t)\right)\left(-\frac{1}{t^2}\right) = 0

Why do you make it so complicated?
Have you learned about "separation of variables"? Collect all terms containing the unknown function at one side of the differential equation and collect all the terms containing "t" on the other side.

u'(t)dt + \frac{1}{t}u(t) = 0

\frac{u'}{u}=-\frac{1}{t}

Integrate both sides

\int {\frac{u'(t)}{u(t)}dt} = -\int {\frac{1}{t}dt}

You can change the integration variable on the left side to u as u'(t)dt = du.

\int {\frac{1}{u}du} = -\int {\frac{1}{t}dt}

\ln(u) = - \ln(|t|) \rightarrow u=1/t.

Now you have to integrate u to get y.

y=\ln(|t| \rightarrow x=t\ln(|t|)

x = t\ln(|t|) is the other linearly independent solution.

ehild

 

[Oxymoron]

That is good new ehild. I actually worked that out just prior to handing in my assignment (why did I make it so complicated? Good question.) but neglected to post back with my findings. Anyway, this is good news because I submitted work similar to yours so I should get it right.