civilnogwu

2022-07-10

The function $y\left(x\right)$ satisfies the linear equation ${y}^{″}+p\left(x\right){y}^{\prime }+q\left(x\right)=0$.
The Wronskian $W\left(x\right)$ of two independant solutions, ${y}_{1}\left(x\right)$ and ${y}_{2}\left(x\right)$ is defined as $|\begin{array}{cc}{y}_{1}& {y}_{2}\\ {y}_{1}^{\prime }& {y}_{2}^{\prime }\end{array}|$. Let ${y}_{1}\left(x\right)$ be given, use the Wronskian to determine a first-order inhomogeneous differential equation for ${y}_{2}\left(x\right)$. Hence show that ${y}_{2}\left(x\right)={y}_{1}\left(x\right){\int }_{{x}_{0}}^{x}\frac{W\left(t\right)}{{y}_{1}\left(t{\right)}^{2}}dt$$\left(\ast \right)$
The only first-order inhomogeneous differential equation I can get for ${y}_{2}\left(x\right)$ is simply ${y}_{1}\left(x\right){y}_{2}^{\prime }\left(x\right)-{y}_{1}^{\prime }\left(x\right){y}_{2}\left(x\right)=W\left(x\right)$. Which is trivial and simply the definition of $W\left(x\right)$, so I don't think this is what it wants. I can easily get to the second result by differentiating $\frac{{y}_{2}\left(x\right)}{{y}_{1}\left(x\right)}$ 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 $x{y}^{″}-\left(1-{x}^{2}\right){y}^{\prime }-\left(1+x\right)y=0$ call this equation $\left(†\right)$, we have already confirmed that ${y}_{1}\left(x\right)=1-x$ is a solution
Hence, using $\left(\ast \right)$ 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 $\left(†\right)$ that is independent of ${y}_{1}$ and satisfies ${y}_{2}\left(0\right)=0,y\prime {\prime }_{2}\left(0\right)=1$.
I'm stuck here too as I don't see how we could expand $W\left(t\right)$ 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 $W\left(x\right)$, I've never worked on these problems before.

soosenhc

Expert

Your arguments sound reasonable to me, and the formulation in the exercise a bit cryptic. But you should perhaps also add that
${W}^{\prime }=|\begin{array}{cc}{y}_{1}& {y}_{2}\\ {y}_{1}^{″}& {y}_{2}^{″}\end{array}|=-pW,$
from which $W\left(x\right)=\mathrm{exp}\left(-P\left(x\right)\right)$ with $P=\int p$. This ensures that in $\left({y}_{2}/{y}_{1}{\right)}^{\prime }=W/{y}_{1}^{2}$ the RHS is just an explicit function of $x$. Might help you on the last bit as well.

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