I was studying the first question of this solution and I don't understand how the professor used integrating factor to solve for the question. My confusion occurs right at the second line of this part:
I do not understand why he is integrating $e^{(\lambda_n^2-r)\tau}$ from $0$ to $t$.
My solution just do indefinite integral and get
$$e^{\lambda_n^2t}v_n = \frac{\sigma_ne^{(\lambda_n^2-r)t}}{\lambda_n^2-r}$$
Someone please help...!
On
From the picture
$\frac{d}{dt}{e^{\lambda^2_n t v_n}}=\sigma_n e^{(\lambda^2_n-r)t}$. Now if you simply integrate LHS and RHS. we get $e^{\lambda^2_n t v_n} = \int_0^t \sigma_n e^{(\lambda^2_n-r)\tau} d\tau +C$. $\tau$ is used because limit is function of $t$. So what is the necessity of the expansion step? I guess.. In the previous lines you may have something like $\dot{v}_n+\lambda_n^2 v_n=\sigma_n e^{-rt}$. Now if you multiply LHS and RHS by $e^{\lambda^2_n t}$, you get the line.
Your indefinite integral result should have a constant $c_n'$ at the end
$$e^{\lambda_n^2t}v_n = \frac{\sigma_ne^{(\lambda_n^2-r)t}}{\lambda_n^2-r}+c_n'$$
where $c_n'$ differs from $c_n$ by
$$c_n' - c_n =- \frac{\sigma_n}{\lambda_n^2-r}$$
The two integral results are equivalent.