# Use the Wronskian to determine a first-order inhomogeneous differential equation for $y_2(x)$.

The function $y(x)$ satisfies the linear equation $y''+p(x)y'+q(x)=0$.

The Wronskian $W(x)$ of two independant solutions, $y_1(x)$ and $y_2(x)$ is defined as $\begin{vmatrix}y_1 & y_2\\ y'_1 & y'_2\\ \end{vmatrix}$. Let $y_1(x)$ be given, use the Wronskian to determine a first-order inhomogeneous differential equation for $y_2(x)$. Hence show that $y_2(x)=y_1(x)\int^x_{x_0}\frac{W(t)}{y_1(t)^2}dt$ $(*)$

The only first-order inhomogeneous differential equation I can get for $y_2(x)$ is simply $y_1(x)y'_2(x)-y'_1(x)y_2(x)=W(x)$. Which is trivial and simply the definition of $W(x)$, so I don't think this is what it wants. I can easily get to the second result by differentiating $\frac{y_2(x)}{y_1(x)}$ and then integrating, but the wording suggests I'm supposed to get to this result from the differential equation that I can't determine.

So I would ask please that someone simply complete the above question, showing the differential equation we are to determine and how to get to the second result from this.

Edit: The question later goes on to give the differential equation $xy''-(1-x^2)y'-(1+x)y=0$ call this equation $(†)$, we have already confirmed that $y_1(x)=1-x$ is a solution

Hence, using $(*)$ with $x_0=0$ and expanding the integrand in powers of $t$ to order $t^3$, find the first three non-zero terms in the power series expansion for a solution, $y_2$, of $(†)$ that is independent of $y_1$ and satisfies $y_2(0)=0, y′′_2(0)=1$.

I'm stuck here too as I don't see how we could expand W(t) as it is surely a function of the unknown $y_2$. Apologies for my ignorance, this is independant study, and aside from a brief look at some online notes that simply define $W(x)$, I've never worked on these problems before.

Thank you,

Your arguments sound reasonable to me, and the formulation in the exercise a bit cryptic. But you should perhaps also add that $$W' = \left| \begin{matrix} y_1 & y_2 \\ y_1'' & y_2 '' \end{matrix} \right| = - pW,$$ from which $W(x)=\exp(-P(x))$ with $P=\int p$. This ensures that in $(y_2/y_1)'=W/y_1^2$ the RHS is just an explicit function of $x$. Might help you on the last bit as well.