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

Sep 5, 2017

$y = A \cos \left(2 \arctan x\right) + B \cos \left(2 \arctan x\right)$

#### Explanation:

${\left(1 + {x}^{2}\right)}^{2} y ' ' + 2 x \left(1 + {x}^{2}\right) y ' + 4 y = 0$ ..... [A]

Perform a substitution

Let $u = \arctan x$

Then:

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$ and $\frac{{d}^{2} u}{{\mathrm{dx}}^{2}} = \frac{- 2 x}{1 + {x}^{2}} ^ 2$

For the first derivative, we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}} \frac{\mathrm{dy}}{\mathrm{du}}$

For the second derivative, we have:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{1}{1 + {x}^{2}} \frac{{d}^{2} y}{{\mathrm{du}}^{2}} \frac{\mathrm{du}}{\mathrm{dx}} + \frac{- 2 x}{1 + {x}^{2}} ^ 2 \frac{\mathrm{dy}}{\mathrm{du}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{1 + {x}^{2}} \frac{{d}^{2} y}{{\mathrm{du}}^{2}} \frac{\mathrm{du}}{\mathrm{dx}} + \frac{- 2 x}{1 + {x}^{2}} ^ 2 \frac{\mathrm{dy}}{\mathrm{du}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{1 + {x}^{2}} ^ 2 \frac{{d}^{2} y}{{\mathrm{du}}^{2}} - \frac{2 x}{1 + {x}^{2}} ^ 2 \frac{\mathrm{dy}}{\mathrm{du}}$

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

${\left(1 + {x}^{2}\right)}^{2} \left\{\frac{1}{1 + {x}^{2}} ^ 2 \frac{{d}^{2} y}{{\mathrm{du}}^{2}} - \frac{2 x}{1 + {x}^{2}} ^ 2 \frac{\mathrm{dy}}{\mathrm{du}}\right\} + 2 x \left(1 + {x}^{2}\right) \left\{\frac{1}{1 + {x}^{2}} \frac{\mathrm{dy}}{\mathrm{du}}\right\} + 4 y = 0$

And we can cancel factors of $\left(1 + {x}^{2}\right)$ giving:

$\frac{{d}^{2} y}{{\mathrm{du}}^{2}} - 2 x \frac{\mathrm{dy}}{\mathrm{du}} + 2 x \frac{\mathrm{dy}}{\mathrm{du}} + 4 y = 0$
$\therefore \frac{{d}^{2} y}{{\mathrm{du}}^{2}} + 4 y = 0$ ..... [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:

$m = \pm 2 i$

Thus the Homogeneous equation [B] has the solution:

${y}_{c} = {e}^{0 x} \left(A \cos \left(2 u\right) + B \cos \left(2 u\right)\right)$
$\setminus \setminus \setminus = A \cos \left(2 u\right) + B \cos \left(2 u\right)$

Now we initially used a change of variable:

$u = \arctan x$

So restoring this change of variable we get:

$y = A \cos \left(2 \arctan x\right) + B \cos \left(2 \arctan x\right)$