Question

Solve the Differential Equation `x^2y'' -3x+5y=0`?

Answer 1

`y=sqrt(x)(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5x`

Explanation:

We have:

`x^2y'' -3x+5y=0`

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 we get:

`x^2((d^2y)/(dt^2)-dy/dt)e^(-2t) -3e^t+5y=0`

`:. (d^2y)/(dt^2)-dy/dt +5y=3e^t` ..... [A]

This is now a second order linear 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-m+5 = 0`

We can solve this quadratic equation, and we get two complex solution:

`m=1/2(1+-sqrt(19)i)`

Thus the Homogeneous equation:

`(d^2y)/(dt^2)-dy/dt +5y=0`

has the solution:

`y=e^(1/2t)(Acos(1/2sqrt(19)t) + Bcos(1/2sqrt(19)t))`

We now seek a Particular Solution of [A], which will be of the form:

`y = Ae^t`

Differentiating wrt `t` we get:

`y' = Ae^t`, and, >`y'' = Ae^t`

Substituting into the DE [A} we get:

`Ae^t-Ae^t+5Ae^t=3e^t`

Equating coefficients we have:

`A-A+5A=3 => A=3/5`

Hence, the GS of [A] is the combination of the homogeneous solution and the particular solution, thus:

`y=e^(1/2t)(Acos(1/2sqrt(19)t) + Bcos(1/2sqrt(19)t)) +3/5e^t`

Now we initially used a change of variable:

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

So restoring this change of variable we get:

`y=e^(1/2lnx)(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5e^lnx`

`:. y=e^(lnsqrt(x))(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5x`

`:. y=sqrt(x)(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5x`

Which is the General Solution