# To find a solution to this Differential Equation?

• Feb 17th 2010, 10:57 AM
Yehia
To find a solution to this Differential Equation?
For the equation:

$\displaystyle dv/dt=-kx^{1/2}$

I do not seem to be able to solve this one to find the general solution! :S
• Feb 17th 2010, 11:43 AM
pickslides
Are you sure that is the equation?

could it be $\displaystyle \frac{dv}{dt}=-kt^{1/2}$

or

$\displaystyle \frac{dv}{dt}=-kv^{1/2}$ ?
• Feb 17th 2010, 12:42 PM
Yehia
Quote:

Originally Posted by pickslides
Are you sure that is the equation?

could it be $\displaystyle \frac{dv}{dt}=-kt^{1/2}$

or

$\displaystyle \frac{dv}{dt}=-kv^{1/2}$ ?

YES sorry my bad! It is the second one! Can you help?
• Feb 17th 2010, 12:46 PM
pickslides
Happy to kick it off for you

$\displaystyle \frac{dv}{dt}=-kv^{1/2}$

separating the variables you get

$\displaystyle \frac{dv}{v^{1/2}}=-k~dt$

Now integrate both sides

Hint: $\displaystyle \int \frac{dv}{v^{1/2}} = 2v^{1/2}$
• Feb 17th 2010, 12:54 PM
Yehia
Quote:

Originally Posted by pickslides
Happy to kick it off for you

$\displaystyle \frac{dv}{dt}=-kv^{1/2}$

separating the variables you get

$\displaystyle \frac{dv}{v^{1/2}}=-k~dt$

Now integrate both sides

Hint: $\displaystyle \int \frac{dv}{v^{1/2}} = 2v^{1/2}$

Right, so therefore the general solution is:

$\displaystyle v=4(-kt+c)^{1/2}$

am i right?
• Feb 17th 2010, 01:43 PM
pickslides
How did you get that?

I get $\displaystyle v = \left( \frac{-kt+C}{2}\right)^2$