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Find the general solution to the exact equation: $$(1+x^2)y''+4xy'+2y = \sin (x) .$$

I understand how to do general solution for first order exact equation, how do we come about to solve second order.

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  • $\begingroup$ See here. $\endgroup$ – Mattos Dec 1 '16 at 7:49
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$u = (1+x^2) y\\ u' = 2x y + (1+x^2)y'\\ u'' = 2y + 4x y' + (1+x^2)y''$

$u'' = \sin x\\ u = -\sin x + Ax + B\\ y = \frac {-\sin x + Ax + B}{1+x^2}$

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  • $\begingroup$ what sort of method is this called $\endgroup$ – ISuckAtMathPleaseHELPME Dec 1 '16 at 15:04
  • $\begingroup$ Haven't a clue what you would call this method. As it happened, the coefficients of $y$ were nicely behaved. If not, it gets a little messier. Nonetheless, you should look for a substitution that alows you to turn a 2nd order diff eq to a first order equation. $\endgroup$ – Doug M Dec 1 '16 at 16:31
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Hint. In simple cases like this (so this is not a general approach), try to see if there is a function $f(x)$ such that $$(f(x)y(x))''=f(x)y''(x)+2f'(x)y'(x)+f''(x)y(x)$$ is equal to the left-hand side of your equation.

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