Linear dependence of solutions
One solution to ty"-(t+2)y'+2y=0 is exp(t)
Find a second linearly independent solution.
https://brainmass.com/math/linear-algebra/linear-dependence-solutions-28803
SOLUTION This solution is FREE courtesy of BrainMass!
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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