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I have this function: $$y''+4y= x-\cos(2x)$$ To proceed I calculated the associated homogeneous, after that I split the original equation into two parts, setting: $u(x) = x$ and $t(x)=-\cos(2x)$ to find two particular solutions. The final solution I've found is: $$y(x)=c_1 \cos(2x)+c_2 \sin(2x)+\frac{1}{4}x -\frac{1}{4}x\sin(2x)$$ Did I any mistake?

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    Did you plug your solution into the equation and check it? By the way, what is sen? – 2017-01-09
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    The homogeneous equation is $y''+4y'=0$. The characteristic polynomial is $\lambda^2+4\lambda=0\implies \lambda=0\vee \lambda=-4$, i.e the homogeneous solution is $y=C_1+C_2e^{-4x}$. Should it be $y''+4y=x-\cos(2x)$? – 2017-01-09
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    Yes you're right the equation is $y''+4y=x-cos(2x)$ – 2017-01-09
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    As Fabian suggested - did you check your solution by plugging it into the equation? Please notice that you can tag users using @username, so they will know you answered. BTW, your answer is correct. – 2017-01-09
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    @Galc127 thanks, but why if i give this equation to wolframealpha the solution that it finds is: $y(x) = c_2 sin(2 x) + c_1 cos(2 x) + x/4 + (sin^2(x))/8 - 1/4 x sin(2 x) - 1/8 cos^2(x)$ – 2017-01-09
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    @andresca, ah, simple. $$\frac{\sin^2(x)-\cos^2(x)}{8}=-\frac{\cos(2x)}{8}$$ and you can redefine $c_1$ to be $c_1-\frac{1}{8}$. – 2017-01-09

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