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Solve the differential equation

$$xy'' + y' = 0$$

using the Green’s function satisfying the initial condition $y(1) = y'(1)$.

Generally, Green's functions are used to solve nonhomogeneous differential equations, where the solution s of the form $$y=Integral(G(x,t)f(t)dt)$$, where G(x,t) is the Green's function and f(t) is the nonhomogeneous term. so if f(t)=0 then wouldn't Green's function method yield a trivial solution? But this question is given as a homework problem to solve. So is there any method to solve it?

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    $\begingroup$ Welcome to MSE. I am not aware of the Green's function, but you can apply the substitution $u = y'$ and solve the corresponding ODE, which is separable and can be solved by the integrating factor as well. $\endgroup$ Commented Feb 27, 2022 at 19:09
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – William
    Commented Feb 27, 2022 at 19:09

2 Answers 2

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As I have said in the comments, make the substitution $u = y'$ in order to get \begin{align*} xu' + u = 0 & \Longleftrightarrow xu' = - u\\\\ & \Longleftrightarrow \frac{u'}{u} = -\frac{1}{x}\\\\ & \Longleftrightarrow \ln|u| = -\ln(x) + c\\\\ & \Longleftrightarrow |u| = \exp(c)\exp(\ln(x^{-1}))\\\\ & \Longleftrightarrow u = \frac{k_{1}}{x}\\\\ & \Longleftrightarrow y' = \frac{k_{1}}{x}\\\\ & \Longleftrightarrow y = k_{1}\ln(x) + k_{2} \end{align*}

where it has been assumed that $x > 0$.

Hopefully this helps !

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Here is an alternative method, when you see terms in $x^py^{(p)}$ where the power of $x$ is the same as the order of derivation for $y$, you are dealing with an Euler-Cauchy ODE.

It is solved by using the substitution $y(x)=u(\ln(x))$.

Here you have : $\ x^2y''+xy'=0\ $ which qualifies and you get $u''(\ln(x))=0$

Therefore $u$ is a polynomial of degree $1$ and $y(x)=a\ln(x)+b$

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