Linear dependence of solutions

Hi there!

Here is the solution in two file formats as usual.
I'm sorry it took me a while, but I wanted to show you two different methods how to approach this kind of problem (who know's in a test it might come handy).

First Method: Wronskian

Write the equation as:

The two solutions must satisfy:

Multiplying equation (2) by y1 and subtracting equation (1) multiplied by y2 we obtain:

Using the Wronskian's definition:

We see that:

So:

This is a first order separable homogenous differential equation and we solve it for W:

Where W0 is just a constant of integration.

So rewriting the equation

As:

We see that

And:

Thus the Wronskian is:

Now, from the definition of the Wronskian we also get:

Which in our case, translates to:

We can solve this integral using integration by parts twice:

Thus (remember that W0 is just an arbitrary constant to be determined by initial conditions so we can absorb the negative sign into the constant):

We can confirm this solution by substituting it back into the equation:

So the solution:

Is the second linearly independent solution to this ODE.

Second method: Reduction of order.

Since we know one solution, we assume a second solution of the form:

Thus:

Substituting this back into the equation:

Since is a solution the term in the last curly parenthesis is zero.

So we are left with:

But here we have only terms of and , so this is actually a first order equation for .

Using the equation becomes:

Substituting we simplify the equation further:

This is a simple separable equation:

But:

Therefore (We calculated this integral before):

And since we have defined we get:

Maple Verification:

> eqn:=t*diff(diff(y(t),t),t)-(t+2)*diff(y(t),t)+2*y(t)=0;

> dsolve(eqn);

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Attachments

• BM 28803.pdf
• BM 28803.doc