(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

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 [tex](x^2 + 1)u\prime\prime + 4xu\prime = 0[/tex]. Reducing the order then by substituting [tex] p = u\prime [/tex] [tex]p\prime = u\prime\prime[/tex] gives us the separable Deq [tex](x^2 +1)\frac{dp}{dx} + 4xp = 0 [/tex].

So

[tex] \frac{1}{p}dp = \frac{-4x}{x^2 +1} dx [/tex]

[tex] ln|p| = -2 ln|x^2 + 1| + C_1 [/tex]

[tex]p = C_1\frac{1}{(x^2 +1)^2} [/tex]

substitute [tex] x = tan \theta[/tex] to integrate the above equation to get U

[tex]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 = [/tex]

[tex]C_1\int cos^2\theta d\theta [/tex]

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

substituting back x for theta we get:

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

and substituting u into the first substitution:

[tex] y = (x^2 +1)(C_1( \frac{tan^{-1} x}{2} + \frac{x}{x^2 +1}) + C_2). [/tex]

Hopefully I've made fewer errors this time around!

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# Homework Help: More reduction of order DiffEq goodness!

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