Question

Solve `(1+x^2)^2y'' + 2x(1+x^2)y'+4y = 0`?

Answer 1

`y = Acos(2arctanx)+Bcos(2arctanx)`

Explanation:

`(1+x^2)^2y'' + 2x(1+x^2)y'+4y = 0` ..... [A]

Perform a substitution

Let `u = arctanx`

Then:

`(du)/dx = 1/(1+x^2)` and `(d^2u)/(dx^2) = (-2x)/(1+x^2)^2`

For the first derivative, we have:

`dy/dx = dy/(du) * (du)/dx = 1/(1+x^2)dy/(du)`

For the second derivative, we have:

`(d^2y)/(dx^2) = 1/(1+x^2)(d^2y)/(du^2) (du)/dx + (-2x)/(1+x^2)^2dy/(du)`
`\ \ \ \ \ \ \ = 1/(1+x^2)(d^2y)/(du^2) (du)/dx + (-2x)/(1+x^2)^2dy/(du)`

`\ \ \ \ \ \ \ = 1/(1+x^2)^2(d^2y)/(du^2) - (2x)/(1+x^2)^2dy/(du)`

Substituting these expressions into the initial Differential Equation [A] we get:

`(1+x^2)^2{1/(1+x^2)^2(d^2y)/(du^2) - (2x)/(1+x^2)^2dy/(du)} + 2x(1+x^2){1/(1+x^2)dy/(du)}+4y = 0`

And we can cancel factors of `(1+x^2)` giving:

`(d^2y)/(du^2) - 2x dy/(du) + 2xdy/(du)+4y = 0`
`:. (d^2y)/(du^2) +4y = 0` ..... [B]

This is a second order linear Homogeneous Differentiation Equation. The standard approach is to find a solution, `y_c` by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives.

Complementary Function

The Auxiliary equation associated with [B] is:

`m^2 + 4 = 0`

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

`m = +-2i`

Thus the Homogeneous equation [B] has the solution:

`y_c = e^(0x)(Acos(2u)+Bcos(2u))`
`\ \ \ = Acos(2u)+Bcos(2u)`

Now we initially used a change of variable:

`u = arctanx`

So restoring this change of variable we get:

`y = Acos(2arctanx)+Bcos(2arctanx)`