Question

Solve the Differential Equation `x^2y'' +11xy'+25y=0`?

Answer 1

`y(x) = c_1x^-5+c_2x^(-5)lnx`

Explanation:

Substitute the variable:

`t=lnx`

`y(x) = phi(lnx) = phi(t)`

so that:

`y'(x) = 1/x phi'`

`y''(x) = 1/x^2 (phi''-phi')`

substituting in the original equation:

`x^2y''+11xy'+25y=0`

`phi''-phi'+11phi'+25phi =0`

`phi''+10phi'+25phi =0`

This is a second order equation with constant coefficients, so we can solve the characteristic equation:

`lambda^2+10lambda+25 =0`

`(lambda+5) = 0`

`lambda = -5`

so the general solution is:

`phi(t) = c_1e^(-5t)+c_2te^(-5t)`

and undoing the substitution:

`y(x) = phi(lnx) = c_1e^(-5lnx)+c_2lnxe^(-5lnx)`

`y(x) = c_1x^-5+c_2x^(-5)lnx`

Answer 2

`y = (Alnx+B)x^(-5)`

Explanation:

We have:

`x^2y'' +11xy'+25y=0` ..... [A]

This is a Euler-Cauchy Equation which is typically solved via a change of variable. Consider the substitution:

`x = e^t => xe^(-t)=1`

Then we have,

`dy/dx = e^(-t)dy/dt`, and, `(d^2y)/(dx^2)=((d^2y)/(dt^2)-dy/dt)e^(-2t)`

Substituting into the initial DE [A] we get:

`x^2((d^2y)/(dt^2)-dy/dt)e^(-2t) +11xe^(-t)dy/dt+25y=0`

`:. ((d^2y)/(dt^2)-dy/dt) +11dy/dt+25y=0`

`:. (d^2y)/(dt^2)+10dy/dt+25y=0` ..... [B]

This is now a second order linear homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

`m^2+10m+25 = 0`

We can solve this quadratic equation, and we get a real repeated root:

`m=-5`

Thus the Homogeneous equation [B] has the solution:

`y=(Ax+B)e^(-5t)`

Now we initially used a change of variable:

`x = e^t => t=lnx`

So restoring this change of variable we get:

`y = (Alnx+B)e^(-5lnx)`

`:. y = (Alnx+B)x^(-5)`

Which is the General Solution.