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I'm trying to solve:

$ y'' -4xy' + (4x^2 -1)y = -3e^{x^2} \sin(2x)$

Which has a general form of

$y'' + P(x) y' + Q(x) = R(x)$

I reduced it to the normal form by using the substitution $y = u e^{ - \frac{1}{2} \int P(x)} dx$

$u'' + I(x) u = S(x)$

Where $I(x) = Q - \frac {p'}{2} - \frac {p^2}{4}$

and $S(x) = R(x) e^{ \frac{1}{2} \int P(x) dx}$

I end up with:

$ u'' + u = -3 \sin(2x)$

I solve first for the complementary function when its homogeneous

$ u_{C.F} = A \cos(x) + B \sin(x)$

Then to solve for the particular solution, I first decided to use variation of parameters, then,

$u_{P.I} = v_1y_1 + v_2y_2$

Where $y1$ and $y2$ are the solutions to the homogeneous equation, I let $y1 = \cos(x)$ and $y2 = \sin(x)$

I compute the Wronskian of both functions since they are linearly independent solutions, $ W(y1(x),y2(x) = \left|\begin{matrix}y1 & y2 \\ y1' & y2' \end{matrix}\right| = \left|\begin{matrix}\cos(x) & \sin(x) \\ - \sin(x) & \cos(x) \end{matrix}\right| = 1$

$ (v_2)' = \frac {-y_2 S(x)}{W} = \frac {- \sin(x) (-3 \sin(2x))}{1} = 6 \sin^2(x) \cos(x)$

$ (v_1)' = \frac {y_1 S(x)}{W} = \frac { \cos(x) (-3 \sin(2x))}{1} = -6 \cos^2(x) \sin(x)$

Integrating $v_1$ and $v_2$ I get,

$v_1 = 2 \cos^3(x)$ , $v_2 = 2sin^3(x)$

Substituting in $u_{P.I} = v_1y_1 + v_2y_2$ ,

$u_{P.I} = 2 [ \cos^4(x) + \sin^4(x)]$

So to get $y_{P.I}$ , $y_{P.I} = u_{P.I} e^{ - \frac{1}{2} \int P(x)} dx$

Therefore, $y_{P.I} = 2e^{x^2} [ \cos^4(x) + \sin^4(x) ]$ <------

This is a solution using variation of parameters, now when I try to do it using undetermined coefficients:

$u'' + u = -3 \sin(2x)$

Let $u_{P.I} = A \sin(2x) + B \cos(2x)$

Skipping some steps, we arrive that $A = 1$ , and $ B = 0$

Then the solution is $u_{P.I} = \sin(2x)

Therefore, $y_{P.I} = e^{x^2} \sin(2x)$ <------

These are two different particular solutions, where's the problem?

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Ok, I got it, when calculating $(v_2)'$ and $(v_1)'$ , they should be:

$(v_2)' = \frac {y_1 S(x)}{W} = \frac { \cos(x) (-3 \sin(2x))}{1} = -6 \cos^2(x) \sin(x)$

$(v_1)' = \frac {-y_2 S(x)}{W} = \frac { -\sin(x) (-3 \sin(2x))}{1} = 6 \sin^2(x) \cos(x)$

Then,

$ v_2 = 2 \ cos^3(x) $ and $v_1 = 2 \sin^3(x)$

Then $u_{P.I} = v_1 y_1 + v_2 y_2 = 2 \sin^3(x) \cos(x) 2 \ cos^3(x) \sin(x) = 2 \sin(x) \cos(x) [ \cos^2(x) + \sin^2(x) ] = 2 \sin(x) \cos(x) = \sin(2x)$

Which is identical to the undetermined coefficients method.

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