- #1
Uku
- 82
- 0
Hello!
On Pauls notes webpage, there is the following problem to be solved by variation of parameters:
[itex]ty''-(t+1)y'+y=t^2[/itex] (1)
On the page, the fundamental set of solutions if formed on the basis of the complementary solution. The set is:
[itex]y_{1}(t)=e^t[/itex] and [itex]y_{2}(t)=t+1[/itex]
Now, I must be missing something here. Since I get the complementary solution for the homogeneous equation of (1):
[itex]r=\frac{(t+1)+/- \sqrt{(t+1)^2-4t}}{2t}[/itex] which solves as [itex]r_{1}=1[/itex] and [itex]r_{2}=\frac{1}{t}[/itex] which would give a complementary solution of:
[itex]Y_{c}=C_{1}e^{t}+C_{2}e^{\frac{1}{t}t}=C_{1}e^{t}+C_{2}e^1[/itex]
from which I would get [itex]y_{1}(t)=e^t[/itex] and [itex]y_{2}(t)=e[/itex]
What have I missed, must be simple...
Regards,
U.
On Pauls notes webpage, there is the following problem to be solved by variation of parameters:
[itex]ty''-(t+1)y'+y=t^2[/itex] (1)
On the page, the fundamental set of solutions if formed on the basis of the complementary solution. The set is:
[itex]y_{1}(t)=e^t[/itex] and [itex]y_{2}(t)=t+1[/itex]
Now, I must be missing something here. Since I get the complementary solution for the homogeneous equation of (1):
[itex]r=\frac{(t+1)+/- \sqrt{(t+1)^2-4t}}{2t}[/itex] which solves as [itex]r_{1}=1[/itex] and [itex]r_{2}=\frac{1}{t}[/itex] which would give a complementary solution of:
[itex]Y_{c}=C_{1}e^{t}+C_{2}e^{\frac{1}{t}t}=C_{1}e^{t}+C_{2}e^1[/itex]
from which I would get [itex]y_{1}(t)=e^t[/itex] and [itex]y_{2}(t)=e[/itex]
What have I missed, must be simple...
Regards,
U.