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Hi I'm really stuck with some homework:

Find the general solution of the differential equation,

$$\left(x+\dfrac{1}x\right)\dfrac{dy}{dx} + 2y = 2\left(x^2+1\right)^2$$

So far, I've divided both sides by $x+\dfrac{1}x$ and integrated $\dfrac{2y}{x + \frac{1}{x}}$ to get $y \ln\left(x^2+1\right)$ but have no idea where to go from here.

Anyone know what I need to do next?

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  • $\begingroup$ Hint: Devide both sides by $(x^2+1)$ and see: math.stackexchange.com/questions/121669/… $\endgroup$
    – NoChance
    Mar 18 '12 at 11:48
  • $\begingroup$ You can't perform that integration. $y$ is not a constant. Have you learned integrating factors? What should you multiply by to get the left side in the form $(uy)'=uy'+u'y$? $\endgroup$
    – Mike
    Mar 18 '12 at 13:43
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Rewrite equation into form :

$$ \frac{dy}{dx}+p(x)y=q(x)$$

General solution is given by :

$$y=\frac{\int u(x)\cdot q(x) \,dx +C}{u(x)} ~\text{where}~u(x)=e^{\int p(x) \,dx}$$

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$$ \begin{align*} (x+(1/x))\frac{dy}{dx} +2y &= 2(x^2+1)^2\\ \left(\frac{x^2+1}{x}\right)\frac{dy}{dx} +2y &= 2(x^2+1)^2\\ \frac{dy}{dx} +\frac{2xy}{x^2+1} &= 2x(x^2+1) \tag{A}\\ \frac{dy}{dx} +P(x)y &= Q(x)\\ \end{align*} $$

where $\displaystyle{P(x) = \frac{2x}{x^2+1}}$ and $\displaystyle{Q(x) = 2x(x^2+1)}$

$$\displaystyle{\int \frac{2x}{x^2+1} = \ln(x^2+1)}$$

The integrating factor = $\displaystyle{e^{\int P(x) dx}}$ which is $\displaystyle{e^{\ln(x^2+1)}} = x^2+1$ (why?)

$(A)$ simplifies to

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

I could finish it completely, but can you figure the rest (by integrating both sides)?

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Your equation can be transformed into a first order linear differential equation. You can find the solving formula here: http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation

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