civilnogwu

Answered

2022-07-10

The function $y(x)$ satisfies the linear equation ${y}^{\u2033}+p(x){y}^{\prime}+q(x)=0$.

The Wronskian $W(x)$ of two independant solutions, ${y}_{1}(x)$ and ${y}_{2}(x)$ is defined as $\left|\begin{array}{cc}{y}_{1}& {y}_{2}\\ {y}_{1}^{\prime}& {y}_{2}^{\prime}\end{array}\right|$. 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}_{0}}^{x}\frac{W(t)}{{y}_{1}(t{)}^{2}}dt$$(\ast )$

The only first-order inhomogeneous differential equation I can get for ${y}_{2}(x)$ is simply ${y}_{1}(x){y}_{2}^{\prime}(x)-{y}_{1}^{\prime}(x){y}_{2}(x)=W(x)$. 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 $x{y}^{\u2033}-(1-{x}^{2}){y}^{\prime}-(1+x)y=0$ call this equation $(\u2020)$, we have already confirmed that ${y}_{1}(x)=1-x$ is a solution

Hence, using $(\ast )$ 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 $(\u2020)$ that is independent of ${y}_{1}$ and satisfies ${y}_{2}(0)=0,y\prime {\prime}_{2}(0)=1$.

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.

The Wronskian $W(x)$ of two independant solutions, ${y}_{1}(x)$ and ${y}_{2}(x)$ is defined as $\left|\begin{array}{cc}{y}_{1}& {y}_{2}\\ {y}_{1}^{\prime}& {y}_{2}^{\prime}\end{array}\right|$. 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}_{0}}^{x}\frac{W(t)}{{y}_{1}(t{)}^{2}}dt$$(\ast )$

The only first-order inhomogeneous differential equation I can get for ${y}_{2}(x)$ is simply ${y}_{1}(x){y}_{2}^{\prime}(x)-{y}_{1}^{\prime}(x){y}_{2}(x)=W(x)$. 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 $x{y}^{\u2033}-(1-{x}^{2}){y}^{\prime}-(1+x)y=0$ call this equation $(\u2020)$, we have already confirmed that ${y}_{1}(x)=1-x$ is a solution

Hence, using $(\ast )$ 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 $(\u2020)$ that is independent of ${y}_{1}$ and satisfies ${y}_{2}(0)=0,y\prime {\prime}_{2}(0)=1$.

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.

Answer & Explanation

soosenhc

Expert

2022-07-11Added 16 answers

Your arguments sound reasonable to me, and the formulation in the exercise a bit cryptic. But you should perhaps also add that

${W}^{\prime}=\left|\begin{array}{cc}{y}_{1}& {y}_{2}\\ {y}_{1}^{\u2033}& {y}_{2}^{\u2033}\end{array}\right|=-pW,$

from which $W(x)=\mathrm{exp}(-P(x))$ with $P=\int p$. This ensures that in $({y}_{2}/{y}_{1}{)}^{\prime}=W/{y}_{1}^{2}$ the RHS is just an explicit function of $x$. Might help you on the last bit as well.

${W}^{\prime}=\left|\begin{array}{cc}{y}_{1}& {y}_{2}\\ {y}_{1}^{\u2033}& {y}_{2}^{\u2033}\end{array}\right|=-pW,$

from which $W(x)=\mathrm{exp}(-P(x))$ with $P=\int p$. This ensures that in $({y}_{2}/{y}_{1}{)}^{\prime}=W/{y}_{1}^{2}$ the RHS is just an explicit function of $x$. Might help you on the last bit as well.

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