On my homework, I was asked to find a second linearly independent solution of the following second order linear DE given the following nontrivial solution $f$:
$$ tx''-(t+1)x'+x=0\quad t>0,f(t)=e^t$$
Here's what I tried:
Let $f=x_1$. Then the second linearly independent solution $x_2=vx_1=ve^t$ for some function $v(t)$.
$x'_2=ve^t+v'e^t$
$\begin{align}
x''_2&=ve^t+v'e^t+v'e^t+v''e^t \\
&=ve^t+2v'e^t+v''e^t
\end{align}$
Substituting $x_2$ into the original equation,
$\begin{align}
t(ve^t+2v'e^t+v''e^t)-(t+1)(ve^t+v'e^t)+ve^t&=0\\
vte^t+2v'te^t+v''te^t-(vte^t+v'te^t+ve^t+v'e^t)+ve^t&=0 \\
vte^t+2v'te^t+v''te^t-vte^t-v'te^t-ve^t-v'e^t+ve^t&=0 \\
v''te^t+v'te^t-v'e^t&=0
\end{align}$
Let $w=v'.$ Then
$\begin{align}
w'te^t+wte^t-we^t&=0 \\
\frac{dw}{dt}te^t&=-wte^t+we^t \\
\int\frac{dw}{w}&=\int\frac{-te^t+e^t}{te^t}dt=\int\left(-1+\frac{1}{t}\right)dt \\
\ln |w|&=-t+\ln t \qquad (t>0) \\
w&=e^{-t+\ln t}=te^{-t}
\end{align}$
$v=\int w\,dt=\int te^{-t}dt=-te^{-t}-e^{-t}$
Finally, $x_2=vx_1=(-te^{-t}-e^{-t})e^t=-t-1.$
The answer in the book was $t+1$, so I have a couple questions:
(1) Is my answer correct? I'm assuming it was since $t+1$ is simply a constant multiple of $-t-1$.
(2) I left out the constants of integration since I was just asked to find a linearly independent function, so the constants of integration seemed to be immaterial because they turn out to be scalar multiples of the function itself. Is this generally how one should go about solving these types of problems, or could I have done it a better way?
Thank you!
Best Answer
(1) Yes! your answer is correct for the reasons you've explained. You can always check if your answer is correct with the original ODE.
(2) Like integrating factors, with reduction of order you can leave out some integration constants. However, if you included them, you'd get that $x_2 = c (t+1)$ and that would automatically check (1).