Question

Find the particular solution of the differential equation that satisfies the initial condition(s).

Answer 1

step 1

we have

$f^{\prime}^{\prime}\left(x\right)=x^{\left(-\frac{3}{2}\right)}$

Find out the integral of the second derivative

$f^{\prime}\left(x\right)=\int x^{\left(-\frac{3}{2}\right)}dx=-\frac{2}{\sqrt{x}}+C$

Find out the value of C

we have

f''(4)=5

$\begin{gathered} 5=-\frac{2}{\sqrt{4}}+C \\ 5=-1+C \\ C=6 \end{gathered}$

therefore

$f^{\prime}\left(x\right)=-\frac{2}{\sqrt{x}}+6$

step 2

Find out the integral of the first derivative

$f\mleft(x\mright)=\int-\frac{2}{\sqrt{x}}+6=-4\sqrt{x}+6x+C$

Find out the value of C

we have

f(0)=0

$\begin{gathered} 0=-4\sqrt{0}+6\left(0\right)+C \\ C=0 \end{gathered}$

therefore

$f\lparen x)=-4\sqrt{x}+6x$