Solving differential equation Below is my work for a particular problem that is mixing me up, since no matter how many times, I can't get my answer to match the book solution.
Given ${f}''(x)= x^{-\frac{3}{2}}$ where $f'(4)= 2$ and $f(0)= 0$, solve the differential equation.
$$f'(x)= \int x^{-\frac{3}{2}} \Rightarrow \frac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1} \Rightarrow -2x^{-\frac{1}{2}} + C$$
$$f'(4)= -2(4)^{-\frac{1}{2}}+ C= 2 \Rightarrow -4+C= 2 \Rightarrow C= 6$$
Thus, the first differential equation is $f'(x)= -2x^{-\frac{1}{2}}+6$
$$f(x)= \int -2x^{-\frac{1}{2}}+6 \Rightarrow 2(\frac{x^{-\frac{1}{2}}}{-\frac{1}{2}+1}) \Rightarrow -4x^{\frac{1}{2}}+6x+C$$
Since $f(0)= 0, C=0$, so the final differential equation should be $f(x)= -4x^{\frac{1}{2}}+6x$, but the book answer has $3x$ in place of my $6x$.  Where did I go wrong?

Nothing to worry about! There is a minor slip, $-2(4)^{-1/2}=-2/2=-1$. You got $-4$ instead. 
