# Thread: Solve DE of the form dx/dt = f(x).

1. ## Solve DE of the form dx/dt = f(x).

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

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

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

2. Originally Posted by isx99
Express your answer in the form $\displaystyle x\ =\ f(t).$
Why don't you do this first, and see if your answer is really wrong?

Subtracting 90 from both sides:
$\displaystyle t\ -\ 90\ =\ -10\sqrt{x\ +\ 1}.$
Therefore:
$\displaystyle x\ +\ 1\ =\ \frac{t^2}{100}\ +\ 81.$
Subtracting 1 from both sides:
$\displaystyle x\ =\ \frac{t^2}{100}\ +\ 80.$

But, when I expand the answer given, I get:
$\displaystyle 9^2\ -\ 2(9)(\frac{1}{10}t)\ +\ \frac{t^2}{100}\ -\ 1.$
Which equals:
$\displaystyle \frac{t^2}{100}\ -\ \frac{9t}{5}\ +\ 80.$

4. Originally Posted by isx99
Subtracting 90 from both sides:
$\displaystyle t\ -\ 90\ =\ -10\sqrt{x\ +\ 1}.$
Therefore:
$\displaystyle x\ +\ 1\ =\ \frac{t^2}{100}\ +\ 81.$
$\displaystyle (a + b)^2 = a^2 + 2ab + b^2$, not $\displaystyle a^2 + b^2$.

-Dan