Question

Solve the following equation by changing the independent variable
(1+x^2)^2 y" +2x(1+x^2) y' + 4y=0

Answer 1

Answer on Question #66660 – Math – Differential Equations

Question

Solve the following equation by changing the independent variable

$$(1 + x ^ {2}) ^ {2} y ^ {\prime \prime} + 2 x (1 + x ^ {2}) y ^ {\prime} + 4 y = 0$$

Solution

We have equation

$$(1 + x ^ {2}) ^ {2} y ^ {\prime \prime} + 2 x (1 + x ^ {2}) y ^ {\prime} + 4 y = 0$$

Divide both sides of the equation by $(1 + x^{2})$

$$(1 + x ^ {2}) y ^ {\prime \prime} + 2 x y ^ {\prime} + \frac {4 y}{(1 + x ^ {2})} = 0$$

For the first and the second terms we have

$$(1 + x ^ {2}) y ^ {\prime \prime} + 2 x y ^ {\prime} = \left((1 + x ^ {2}) y ^ {\prime}\right) ^ {\prime}$$

The equation takes the form

$$\left((1 + x ^ {2}) y ^ {\prime}\right) ^ {\prime} + \frac {4 y}{(1 + x ^ {2})} = 0$$

or

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

Change the independent variable $x$ by $t = \arctan x$. Replace the variable in the derivative

$$\frac {d}{d x} = \frac {d}{d t} \frac {d t}{d x} = \frac {d}{d t} \frac {d (\arctan x)}{d x} = \frac {d}{d t} \cdot \frac {1}{1 + x ^ {2}}$$

Multiplying both sides of equality by $1 + x^{2}$ we get

$$(1 + x ^ {2}) \frac {d}{d x} = \frac {d}{d t}$$

The equation takes the form

$$\frac {d ^ {2} y}{d t ^ {2}} + 4 y = 0$$

The solution of this equation is

$$y = A \cos 2 t + B \sin 2 t$$

where $A$ and $B$ are arbitrary real constants.

Now we change the variable $t$ by $x$ using the formula

$$\cos 2 t = \frac {1 - \tan^ {2} t}{1 + \tan^ {2} t}, \quad \sin 2 t = \frac {2 \tan t}{1 + \tan^ {2} t}$$

Since $x = \tan t$ we get

$$\cos 2 t = \frac {1 - x ^ {2}}{1 + x ^ {2}}, \quad \sin 2 t = \frac {2 x}{1 + x ^ {2}}$$

Then the solution of the original equation is

$$y = A \frac {1 - x ^ {2}}{1 + x ^ {2}} + B \frac {2 x}{1 + x ^ {2}}$$

Answer: $y = A\frac{1 - x^2}{1 + x^2} + B\frac{2x}{1 + x^2}$

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