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

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?

• 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. Commented Feb 27, 2022 at 19:09
• 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. Commented Feb 27, 2022 at 19:09

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 !

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