# If {f}''\left ( x \right )= 3x^2-5\sin x+ 2\varrho^x, {f}'\left(0\right)=10 and...

## Question:

If {eq}{f}''\left ( x \right )= 3x^2-5\sin x+ 2\varrho^x, {f}'\left(0\right)=10 and f\left(0\right)=0 {/eq}, then what is f(x)?

## Initial Value Problem:

Note that in the problem above we are provided with the second derivative of a function, and in addition we are also provided with two initial values. These kinds of problems are called initial value problems, IVP for short. In order to solve this problem, we will need to integrate a couple times, and apply the initial conditions to find the constants of integration that we will pick up along the way.

First, let's rewrite the exponential function as a natural exponential function.

{eq}\begin{align*} 2 \rho^x &= 2 e^{\ln \rho^x}\\ &= 2 e^{(\ln \rho)\ x} \end{align*} {/eq}

We antidifferentiate divine to find

{eq}\begin{align*} f' (x) &= \int 3x^2-5\sin x+ 2 e^{(\ln \rho)\ x} \ dx \\ &= x^3 + 5 \cos x + \frac2{\ln \rho}\ e^{(\ln \rho)\ x} + C \end{align*} {/eq}

Then, since {eq}f'(0)=10 {/eq} we have

{eq}\begin{align*} 0 + 5 \cos 0 + \frac2{\ln \rho}\ e^{0} + C &= 10 \\ C &= 5 - \frac2{\ln \rho} \end{align*} {/eq}

And so

{eq}\begin{align*} f' (x) &= x^3 + 5 \cos x + \frac2{\ln \rho}\ e^{(\ln \rho)\ x} +5 - \frac2{\ln \rho} \end{align*} {/eq}

We antidifferentiate again to find

{eq}\begin{align*} f (x) &= \int x^3 + 5 \cos x + \frac2{\ln \rho}\ e^{(\ln \rho)\ x} +5 - \frac2{\ln \rho} \ dx\\ &= \frac14 x^4 + 5\sin x + \frac2{\ln^2 \rho}\ e^{(\ln \rho)\ x} +\left (5 - \frac2{\ln \rho} \right ) x + D \end{align*} {/eq}

And since {eq}f(0) = 0 {/eq} we have

{eq}\begin{align*} 0+ 5\sin 0 + \frac2{\ln^2 \rho}\ e^{0} +0 + D &= 0\\ D &= - \frac2{\ln^2 \rho} \end{align*} {/eq}

So we ultimately find

{eq}\begin{align*} f (x) &= \boldsymbol{ \frac14 x^4 + 5\sin x + \frac2{\ln^2 \rho}\ \rho^x +\left (5 - \frac2{\ln \rho} \right ) x - \frac2{\ln^2 \rho} } \end{align*} {/eq}