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Given the second-order ordinary differential equation: $$ {y}''+y=f(x) $$ prove that: $$ y_p(x)=\int_{0}^{x}f(u)\sin(x-u)du $$ is the particular solution of the equation.

I know this is homework but I've been trying to solve it for the past few days and I can't. I even asked my teacher for help but he doesn't answer.

Thanks.

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  • $\begingroup$ try to calculate $y_p''$ using FCT $\endgroup$ – Martín Vacas Vignolo Dec 30 '17 at 1:02
  • $\begingroup$ How is this a PDE? There's only one independent variable here $\endgroup$ – Dylan Dec 30 '17 at 1:08
  • $\begingroup$ Try to read the differential equation mentally: it says "the double derivative of $y$ plus $y$ itself will give us $f(x)$". So, when we need to check if a given $y_P$ is a solution to the deq or not, we need to calculate its double derivative first. $\endgroup$ – Gaurang Tandon Dec 30 '17 at 1:26
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Hint. We assume our $f$ is nice enough to be allowed to use the Leibniz rule, $$ \frac{d}{dx} \left (\int_{0}^{b(x)}f(x,u)\,du \right) = f\big(x,b(x)\big)\cdot \frac{d}{dx} b(x) + \int_{0}^{b(x)}\frac{\partial}{\partial x} f(x,u) \,du $$ giving here $$ \begin{align} y'_P(x)&=\frac{d}{dx} \left (\int_{0}^{x}f(u)\sin(x-u)\,du \right) \\\\&= 0+ \int_{0}^{x}f(u) \cos(x-u) \,dt \end{align} $$ then differentiate once more using the same tool and see what happens.

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Variation of constants tells you to find $y_p$ in the form $$ y_p(x)=c_1(x)\cos(x)+c_2(x)\sin(x) $$ where the coefficient functions satisfy the differential equation system \begin{align} c_1'(x)\cos x+c_2'(x)\sin x &= 0\\ -c_1'(x)\sin x+c_2'(x)\cos x &= f(x) \end{align} which leads to $$ c_1'(x)=-f(x)\sin x,\ \ c_2'(x)=f(x)\cos(x) $$ and thus $$ y_p(x)=-\int_0^xf(u)\sin u\,du\,\cos x+\int_0^xf(u)\cos u\,du\,\sin x =\int_0^xf(u)\sin(x-u)\,du. $$

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