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

##### 1 Answer

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

#### Explanation:

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

# (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,

**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) #