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