# Homework Help: More reduction of order DiffEq goodness!

1. May 24, 2010

### bitrex

1. The problem statement, all variables and given/known data

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

3. 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!

Last edited: May 24, 2010
2. May 24, 2010

### gabbagabbahey

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

3. May 24, 2010

### 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!

4. May 24, 2010

### 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.