How to solve homogeneous differential equation with initial value conditions using Green's function?

Question

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?

Answer 1

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 !

Answer 2

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 $y(x)=a\ln(x)+b$