The Question:Solve the differential equation:

$\displaystyle \frac{dx}{dt}\ =\ -\frac{1}{5}(x\ +\ 1)^{\frac{1}{2}}$, given that $\displaystyle x\ =\ 80$ when $\displaystyle t\ =\ 0.$

Express your answer in the form $\displaystyle x\ =\ f(t).$

My Attempt:Integrate both sides:

$\displaystyle \int\frac{dx}{dt}\ =\ \int{-\frac{1}{5}(x+1)^{\frac{1}{2}}$.

Therefore:

$\displaystyle \int\frac{dx}{\sqrt{x\ +\ 1}}\ dx\ =\ \int{-\frac{1}{5}}$.

So:

$\displaystyle \int\frac{1}{\sqrt{x+1}}\ dx\ =\ -\frac{1}{5}t$.

Let $\displaystyle u\ =\ x\ +\ 1$.

$\displaystyle \frac{du}{dx}\ =\ 1$, so $\displaystyle du\ =\ dx.$

Substituting:

$\displaystyle \int{\frac{1}{\sqrt{u}}}}\ du\ =\ -\frac{1}{5}t.$

Simplifying:

$\displaystyle 2\sqrt{x\ +\ 1}\ +\ C\ =\ -\frac{1}{5}t$.

$\displaystyle -t\ =\ 10\sqrt{x\ +\ 1}\ +\ 5C.$

$\displaystyle t\ =\ -10\sqrt{x\ +\ 1}\ -\ 5C.$.

Substitute $\displaystyle x\ =\ 80$, $\displaystyle t\ =\ 0:$

$\displaystyle 0\ =\ -10(9)\ -5C$, so $\displaystyle C\ =\ 18.$

Therefore:

$\displaystyle t\ =\ -10\sqrt{x\ +\ 1}\ +\ 90.$.

However, the answer given is:

$\displaystyle x\ =\ (9\ -\ \frac{1}{10}t)^{2}\ -\ 1.$

Please provide a hint as to where I may have gone wrong.