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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?

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+1 for showing your work. That said, the arrows are not very informative; "$\Rightarrow$" is usually interpreted to mean "implies". The fact that $f'(x)=\int x^{-3/2}\,dx$ (remember the $dx$) does not imply the next formula, which does not imply the next formula, etc. – Arturo Magidin Apr 14 '11 at 20:43
up vote 5 down vote accepted

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

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that was it. I missed the negative sign and interpreted it as $-2\sqrt{4}$ instead of $-\frac{2}{\sqrt{4}}$. Thanks! – Jason Apr 15 '11 at 0:55

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