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If $$\phi(x)=\cos(x)-\int_0^x(x-t)\phi(t)dt$$ then what is the value of $\phi(x)+\phi''(x)$ ?

My Attempt:

$$\phi'(x)=-\sin(x)-(x-x)\phi(x)\dfrac{d(x)}{dx}+(x-0)\dfrac{d(0)}{dx}=-\sin(x)$$ (using Leibniz Rule)

$$\implies \phi''(x)=-\cos(x)$$

So, $$\phi(x)+\phi''(x)=-\int_0^x(x-t)\phi(t)dt$$

How to simplify this? The final answer is given as $-\cos(x)$ but I am not getting it. Please help.

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    $$\phi'(x)=-\sin(x)-\int_0^x\frac{\partial }{\partial x}((x-t)\phi(t))dt=-\sin x-\int_0^t \phi(t) dt $$ – 2017-01-06
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    @Rohan What do you say? Please clarify your sentence. – 2017-01-06
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    @Behrouz Maleki All clear. Couldn't get it at first, thats all. Also the integral is from 0 to x by the way. – 2017-01-06
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    @BehrouzMaleki Thank you. I got it now :)! – 2017-01-06
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    @Rohan It was a typo. Your previous sentence (that you delete it ) was completely wrong . – 2017-01-06
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    @Rohan See it https://en.wikipedia.org/wiki/Leibniz_integral_rule – 2017-01-06
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    @Behrouz Maleki Yes I blabered something without thinking and looking properly at your comment . Sorry. – 2017-01-06

2 Answers 2

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$\displaystyle \phi (x) = \cos x- \int^{x}_{0}(x-t)\phi(t)dt= \cos x-\int^{x}_{0}x\phi(t)dt+\int^{x}_{0}t\phi(t)dt$

then $\displaystyle \phi'(x) = -\sin x-x\phi(x)-\int^{x}_{0}\phi (t)dt+x\phi(x)$

and $\displaystyle \phi''(x) = -\cos x-x\phi'(x)-\phi (x)-\phi (x)+x \phi'(x)+\phi(x)$

So $\phi''(x) = -\cos x-\phi(x)$

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    It is exactly what I wrote. – 2017-01-06
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    Basically the same as below but for a small typo. Error corrected. Not the down voter (in case you are wondering?) – 2017-01-06
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    yes i have typed very slow that,s why you have beat me – 2017-01-06
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Note that $$\phi\left(x\right)=\cos\left(x\right)-x\int_{0}^{x}\phi\left(t\right)dt+\int_{0}^{x}t\phi\left(t\right)dt. $$ Using the fundamental theorem of calculus we have $$\phi'\left(x\right)=-\sin\left(x\right)-\int_{0}^{x}\phi\left(t\right)dt $$ $$\phi''\left(x\right)=-\cos\left(x\right)-\phi\left(x\right) $$ hence $$\phi''\left(x\right)+\phi\left(x\right)=-\cos\left(x\right).$$