Question

What is the general solution of the differential equation `x^2y'' -xy'-3y=0`?

Answer 1

see Below

Explanation:

There is a certain way to find the general solution if a one solution is already given. It is said that if given solution `y_1`, the second one is given as `y_2=v(x)y_1`. Where in this problem, `y_1=1/x`

If `y_2=v(x)1/x`, then the derivatives are:
`y'_2=v(-1/(x^2))+v'(1/x)`

and

`y''_2=v(2/x^3)+v'(-2/x^2)+v''(1/x)`

Substitute this values in the original equation:

`x^2y''-xy'-3y=0`

And:

`x^2(v(2/x^3)+v'(-2/x^2)+v''(1/x))-x(v(-1/(x^2))+v'(1/x))-3(v(1/x))=0`

Although this is suprisingly long, this actually simplifies to:

`v''-3/xv'=0`

Substitute `w=v'` so that is easy to solve as first order differential equation.

`w'-3/xw=0`

The integrating factor is `1/x^3`(Given by `e^(int3/xdx)`).

`1/x^3w'-3/x^4w=0`
`(1/x^3w)'=0`

Integrate both sides:

`1/x^3w=C_1`

Undo substitution:

`v'1/x^3=C_1`
`v'=C_1x^3`

Integrate both sides again:

`v=C_1/4x^4+C_2`

Now the general solution is given as:
`y_2=v(x)y_1`
`y_2=(C_1/4x^4+C_2)(1/x)`
`y_2=(C_1(x^3/4)+C_2(1/x))`

Now the first solution is given when `C_1=0`, then the second solution can be when `C_2=0` which is `x^3/4` otherwise there are infinite solutionsby just filling the constants.

To check:
`y=x^3/4`
`y'=3/4x^2`
`y''=3/2x`

Substitute:
`x^2(3/2x)-x(3/4x^2)-3(x^3/4)=0`
`3/2x^3-3/4x^3-3/4x^3=0`
`6x^3-3x^3-3x^3=0`
`0=0`

Which is correct.

Answer 2

`y=A/x + Bx^3`

Explanation:

We have:

`x^2y'' -xy'-3y=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) -xe^(-t)dy/dt-3y=0`

`:. ((d^2y)/(dt^2)-dy/dt) -dy/dt-3y=0`

`:. (d^2y)/(dt^2)-2dy/dt-3y=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^2-2m-3 = 0`

We can solve this quadratic equation, and we get two real and distinct solutions:

`(m+1)(m-3) = 0 => m=-1,3`

Thus the Homogeneous equation [B]:

`:. (d^2y)/(dt^2)-2dy/dt-3y=0`

has the solution:

`y=Ae^(-t) + Be^(3t)`

Now we initially used a change of variable:

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

So restoring this change of variable we get:

`y=Ae^(-lnx) + Be^(3lnx)`

`:. y=Ae^(lnx^(-1)) + Be^(lnx^3)`

`:. y=Ax^(-1) + Bx^3`

`:. y=A/x + Bx^3`

Which is the General Solution