Problem with finding the complementary solution of ODE

[Uku]

Hello!

On Pauls notes webpage, there is the following problem to be solved by variation of parameters:

$ty''-(t+1)y'+y=t^2$ (1)
On the page, the fundamental set of solutions if formed on the basis of the complementary solution. The set is:
$y_{1}(t)=e^t$ and $y_{2}(t)=t+1$

Now, I must be missing something here. Since I get the complementary solution for the homogeneous equation of (1):

$r=\frac{(t+1)+/- \sqrt{(t+1)^2-4t}}{2t}$ which solves as $r_{1}=1$ and $r_{2}=\frac{1}{t}$ which would give a complementary solution of:

$Y_{c}=C_{1}e^{t}+C_{2}e^{\frac{1}{t}t}=C_{1}e^{t}+C_{2}e^1$
from which I would get $y_{1}(t)=e^t$ and $y_{2}(t)=e$

What have I missed, must be simple...

Regards,
U.

 

[tiny-tim]

Hello Uku!

I get the complementary solution for the homogeneous equation of (1):

$r=\frac{(t+1)+/- \sqrt{(t+1)^2-4t}}{2t}$ which solves as $r_{1}=1$ and $r_{2}=\frac{1}{t}$ which would give a complementary solution of:

$Y_{c}=C_{1}e^{t}+C_{2}e^{\frac{1}{t}t}$ …

no, the characteristic polynomial method only works for constant coefficients,

not for coefficients which depend on t

 

[Uku]

Okay, that is true, thank you. I now read from his example that the set is given by default.

Still: how would you arrive at $y_{1}$ and $y_{2}$?

U.

 

[tiny-tim]

dunno

 

[blue_raver22]

Hello U.,

It seems like you have made a mistake in your calculation for the complementary solution. The correct solution should be y_{c}=C_{1}e^{t}+C_{2}(t+1). This can be verified by plugging it into the original differential equation and solving for the constants.

In general, finding the complementary solution for a differential equation can be tricky and often requires careful calculations and attention to detail. It is important to double check your work and make sure all steps are correct. If you are still having trouble, I would recommend seeking help from a colleague or a tutor to ensure accuracy in your solution. Good luck!