Find differential equation

Question

Given the differential equation $y''+a(t)y'+b(t)y=0$, it has a solution $y(t) = t$ for $t>0$. Suppose that the Wronskian associated with two linearly independent solutions has the form $W(t) = 1-\ln(t)$. Determine the functions $a(t)$ and $b(t)$ explicitly.

My steps:

1. After hours of work, I found the second soluction is $y_2(t) = \ln(t)$

My question is any idea to use this information, to find $a(t)$ and $b(t)$.

Thanks

Answer 1

Since you didn't specify the method you utilized in order to find that second solution, i'm going to inform the whole process i thought:

First instinct is to compute the Wronskian using the solution you already have to try and find the second solution. This will give rise to another differential equation:

$$t\frac{dy_2(t)}{dt}-y_2=1-\ln(t)$$

Which constitutes a first order homogeneous differential equation. You can solve for $y_2(t)$ (divide the entire equation by $t$ and try to force the product rule of derivatives on the left hand side using a function $u(t)$). Then, with the second solution at hand, substitute both solutions in the original differential equation and find $a(t)$ and $b(t)$.

Answer 2

There are basically two ways to solve this problem.

The first method, which you start to follow, needs to find the second solution to the differential equation thanks to the given Wronskian. Indeed, since $W = y_1y_2' - y_1'y_2 = 1 - \ln t$, with $y_1(t) = t$, we end up with an inhomogeneous linear first-order differential equation for $y_2(t)$, namely $ty_2' - y_2 = 1 - \ln t$. Applying the method of variation of the parameter to this last equation, we get $y_2(t) = \ln t$, which you have determined already.

Now, since $y_1$ and $y_2$ both satisfy the original differential equation, we can write two equations for the unknown functions $a(t)$ and $b(t)$, i.e. $$ \begin{cases} y_1'' + ay_1' + by_1 = 0 \\ y_2'' + ay_2' + by_2 = 0 \end{cases} $$ Those equations are linear, this is why we can recast them as the following matrix equation : $$ \begin{pmatrix} y_1' & y_1 \\ y_2' & y_2 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = - \begin{pmatrix} y_1'' \\ y_2'' \end{pmatrix} $$ or, more explicitly, $$ \begin{pmatrix} 1 & t \\ 1/t & \ln t \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 1/t^2 \end{pmatrix} $$ I let you finish and invert this equation.

The second method is more direct and doesn't require a second solution. It is based on the fact that the Wronskian satisfy $W' = -aW$, hence $a = -\frac{W'}{W} = \frac{1}{t(1-\ln t)}$ in the present case. In consequence, we have straightforwardly $b = -\frac{y_1''+ay_1'}{y_1} = -\frac{1}{t^2(1-\ln t)}$.