Further Thoughts on Variation of Parameters

Most of the problems that I have you look at involving variation of parameters could also be solved using undetermined coefficients. Variation of paramters extends to classes of differential equations which undetermined coefficients does not, so it is a useful technique. However, it is easy to run into integrals which are difficult to do, and it is not easy to spot problems which this is likely to arise without solving them

For example, consider the differential equation:

L[y_]:=4D[y,{t,2}]+y==Sec[t/2]

L[y[t]]

Undetermined coefficients will not work (g(t) is not one of the special forms), so we are forced to try variation of parameters. But we don't know how much work it will involve. Let's use Mathematica to help us guess how "difficult" the task of solving this by hand using variation of parameters would be.

sol=DSolve[L[y[t]],y[t],t]

This solution looks a bit long, but not terribly so--variation of parameters can probably be accomplished by hand without too much difficulty.

Let's do the same thing for the following differential equation:

L[y_]:=4D[y,{t,2}]+y==Sec[t]

L[y[t]]

If anything, this one looks "simpler" in some sense than the previous differential equation. So we might think it would be about as much work to solve. Let's get Mathematica to solve it for us:

sol=DSolve[L[y[t]],y[t],t]

This solution looks far more complicated than the previous one! Youch! So you can't tell by looking at a differential equation how much work variation of parameters might be. The difference between these two differential equations is in the integrals which have to be done in variation of parameters--and we all know that doing integrals is a difficult task in general.

Here's another differential equation with a similar g(t). Mathematica has great difficulty solving this. I got sick of waiting, so I don't know if Mathematica can solve it!

L[y_]:=4D[y,{t,2}]+y==Sec[2t]
L[y[t]]
sol=DSolve[L[y[t]],y[t],t]

Example 3.6.1

Check the solution of the differential equation:

sol=DSolve[{y''[t]-5y'[t]+6y[t]==2Exp[t]},y[t],t]

Example 3.6.4

Check the solution of the differential equation:

sol=DSolve[{4y''[t]-4y'[t]+y[t]==16Exp[t/2]},y[t],t]

Example 3.6.13

Check the solution of the differential equation:

sol=DSolve[{t^2y''[t]-2y[t]==3t^2-1},y[t],t]