Let D be the domain of a twice differentiable | vedclass

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(C) Given $f^{\prime \prime}(x) + f(x) = 0$. Since $f(x) = \int g(x) \, dx + C$,by differentiating both sides,we get $f^{\prime}(x) = g(x)$. Now,subst

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(C) Given $f^{\prime \prime}(x) + f(x) = 0$. Since $f(x) = \int g(x) \, dx + C$,by differentiating both sides,we get $f^{\prime}(x) = g(x)$. Now,substitute $g(x) = f^{\prime}(x)$ into the expression for $h(x)$: $h(x) = {f(x)}^2 + {f^{\prime}(x)}^2$. To find the rate of change of $h(x)$,differentiate with respect to $x$: $h^{\prime}(x) = 2f(x)f^{\prime}(x) + 2f^{\prime}(x)f^{\prime \prime}(x)$. $h^{\prime}(x) = 2f^{\prime}(x) [f(x) + f^{\prime \prime}(x)]$. Since $f^{\prime \prime}(x) + f(x) = 0$,we have $h^{\prime}(x) = 2f^{\prime}(x) \cdot 0 = 0$. Since the derivative of $h(x)$ is $0$,$h(x)$ is a constant function. Therefore,$h(2015) = h(2014) = h(0) = 5$. Thus,$h(2015) - h(2014) = 5 - 5 = 0$.

Let $D$ be the domain of a twice differentiable function $f$. For all $x \in D, f^{\prime \prime}(x)+f(x)=0$ and $f(x)=\int g(x) \, dx + \text{constant}$. If $h(x)={f(x)}^2+{g(x)}^2$ and $h(0)=5$,then $h(2015)-h(2014)$ is equal to

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