# Differential Equation

1. Mar 22, 2005

### 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??

2. Mar 23, 2005

### ehild

You missed a "t".

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

ehild

3. Mar 23, 2005

### Oxymoron

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

4. Mar 23, 2005

### 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?

Last edited: Mar 23, 2005
5. Mar 23, 2005

### dextercioby

Make the substitution

$$y'(t)=u(t)$$

Daniel.

6. Mar 23, 2005

### 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?

7. Mar 29, 2005

### ehild

Why do you make it so complicated?
Have you learnt 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

8. Mar 29, 2005

### 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.