# Verify that the indicated function is a solution of the given differential equation. Assume an...

## Question:

Verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval {eq}I {/eq} of definition of the solution.

{eq}y''+y=e^{x^2}; \ y=\sin x \int _0^x e^{t^2} \cos t \ dt - \cos x \int ^x_0 e^{t^2} \sin t \ dt {/eq}

## Fundamental Theorem of Calculus:

Given a continuous function, we can build a differentiable function from the concept of integral, that is, a function with variable upper limit, following the expression:

{eq}g\left( x \right) = \int_a^x {f\left( t \right)} dt \to g'\left( x \right) = f\left( x \right) {/eq}.

## Answer and Explanation:

First, using the product rule and the Fundamental Theorem of Calculus, we can calculate the first derivative of the function:

{eq}y'' + y = {e^{{x^2}}};\;y = \sin x\int_0^x {{e^{{t^2}}}} \cos t\;dt - \cos x\int_0^x {{e^{{t^2}}}} \sin t\;dt\\ \\ y' = \cos x\int_0^x {{e^{{t^2}}}} \cos t\;dt + \sin x{e^{{x^2}}}\cos x + \sin x\int_0^x {{e^{{t^2}}}} \sin t\;dt - \cos x{e^{{x^2}}}\sin x\\ y' = \cos x\int_0^x {{e^{{t^2}}}} \cos t\;dt + \sin x\int_0^x {{e^{{t^2}}}} \sin t\;dt {/eq}

In the same way, the second derivative of the function can be written as:

{eq}y'' = - \sin x\int_0^x {{e^{{t^2}}}} \cos t\;dt + \cos x{e^{{x^2}}}\cos x + \cos x\int_0^x {{e^{{t^2}}}} \sin t\;dt + \sin x{e^{{x^2}}}\sin x\\ y'' = - \sin x\int_0^x {{e^{{t^2}}}} \cos t\;dt + {e^{{x^2}}}\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + \cos x\int_0^x {{e^{{t^2}}}} \sin t\;dt\\ y'' = - \sin x\int_0^x {{e^{{t^2}}}} \cos t\;dt + {e^{{x^2}}} + \cos x\int_0^x {{e^{{t^2}}}} \sin t\;dt {/eq}

Now, we can prove this function is a solution of the differential equation:

{eq}y'' + y = - \sin x\int_0^x {{e^{{t^2}}}} \cos t\;dt + {e^{{x^2}}} + \cos x\int_0^x {{e^{{t^2}}}} \sin t\;dt + \sin x\int_0^x {{e^{{t^2}}}} \cos t\;dt - \cos x\int_0^x {{e^{{t^2}}}} \sin t\;dt\\ y'' + y = {e^{{x^2}}} {/eq}

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