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I'd really love your help with the following problem. I need to make linearization and then to find a general solution of the accepted equation for the differential equation:

$y''+\sin(2x+\cos(2y'+y))+1-\sin(y+3y')=0$

I usually get a system of two equations, derive them and go on. What should I do in this case? is there a way to turn it into a system of two equation of order one? How it's been done?

Thanks a lot

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The easiest linearization I can think of in this case is to use $\cos(\theta) \approx 1$, and $\sin(\theta) \approx \theta$, where $\theta$ is small enough. In your case, these assumptions boil down to making the assumption that $y,y' \approx 0$.

With these assumptions, your differential equation now becomes $y'' + \sin(2x+1) + 1 - (y+3y') = 0$ which in turn becomes $y'' - 3y' - y = - \left(1 + \sin(2x+1) \right)$ Now you can use your differential equation tricks to solve the above equation to get $y(x) = c_1 \exp \left( \left(\dfrac{3-\sqrt{13}}2 \right) x\right) + c_2 \exp \left( \left(\dfrac{3+\sqrt{13}}2 \right) x\right) + \dfrac{5 \sin(2x+1) - 6 \cos(2x+1) + 61}{61}$

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    Set $z = y'$ to get two equations $z' - 3z - y = - (1+\sin(2x+1))$ and $z=y'$2012-06-30