More reduction of order DiffEq goodness

[bitrex]

Homework Statement

Use the substitution y = (x^2 + 1)u to solve the differential equation (x^2 +1)y\prime\prime = 2y

The Attempt at a Solution

I was having some trouble with these earlier because I needed to brush up on my trigonometric substitution. Let's try this one...

Making the substitution and simplifying the resultant equation gives us the differential equation (x^2 + 1)u\prime\prime + 4xu\prime = 0. Reducing the order then by substituting p = u\prime p\prime = u\prime\prime gives us the separable Deq (x^2 +1)\frac{dp}{dx} + 4xp = 0.

So

\frac{1}{p}dp = \frac{-4x}{x^2 +1} dx

ln|p| = -2 ln|x^2 + 1| + C_1

p = C_1\frac{1}{(x^2 +1)^2}

substitute x = tan \theta to integrate the above equation to get U

u = C_1\int\frac{ sec^2\theta}{(1+tan^2\theta)^2}d\theta = C_1\int\frac{1}{sec^2\theta}d\theta =

C_1\int cos^2\theta d\theta

C_1\int \frac{1}{2} + \frac{cos2\theta}{2}d\theta = C_1(\frac{\theta}{2} + sin\theta cos\theta + C_2)

substituting back x for theta we get:

u = C_1( \frac{tan^{-1} x}{2} + \frac{x}{x^2 +1} + C_2)

and substituting u into the first substitution:

y = (x^2 +1)(C_1( \frac{tan^{-1} x}{2} + \frac{x}{x^2 +1}) + C_2).Hopefully I've made fewer errors this time around!

 

[gabbagabbahey]

\cos^2\theta\neq\frac{1}{2}[1+\sin(2\theta)]

 

[bitrex]

I don't see where I made that substitution? I now see I did make an error on this line:C_1\int \frac{1}{2} + \frac{cos2\theta}{2}d\theta = C_1(\frac{\theta}{2} + sin\theta cos\theta + C_2)

The integral of \frac{cos(2\theta)}{2} = \frac{sin(2\theta)}{4}, which I then replaced with a trigonometric identity should end up as \frac{sin(\theta)cos(\theta)}{2}...fortunately in this case I don't think the damage is too great as the factor of 1/2 should be swallowed by the constant!

 

[gabbagabbahey]

The factor of 1/2 isn't swallowed by the constant since the same constant multiplies another term. Fix the rest of your calculation with the 1/2 and you should be fine.