# Thread: first order differential question

1. ## first order differential question

hi i am studying some past papers for my exam at uni and would like help with this question if it is not a problem....
y= Ae^-ax show that dy/dx + ay = 0

this would be very much appreciated!

2. Originally Posted by Ep!taph
hi i am studying some past papers for my exam at uni and would like help with this question if it is not a problem....
y= Ae^-ax show that dy/dx + ay = 0

this would be very much appreciated!
All you need to do is verify that the differential equation is true.

Since $\displaystyle y=Ae^{-ax}$, it follows that $\displaystyle \frac{\,dy}{\,dx}=-aAe^{-ax}$.

Now,

$\displaystyle \frac{\,dy}{\,dx}+ay=-aAe^{-ax}+a\left(Ae^{-ax}\right)=aAe^{-ax}-aAe^{-ax}=0$.

Therefore, we showed that $\displaystyle y=Ae^{-ax}$ satisfied the differential equation $\displaystyle \frac{\,dy}{\,dx}+ay=0$.

Does this make sense?

dy/dx + 3y = 6
By the method of separating the variables determine the equation that describes 'y' given that y = 0 when x =0

4. Originally Posted by Ep!taph

dy/dx + 3y = 6
By the method of separating the variables determine the equation that describes 'y' given that y = 0 when x =0
dy/dx= 6- 3y= 3(2-y)
"Separating the variables" means writing it as
$\displaystyle \frac{dy}{2- y}= 3 dx$

Can you integrate those?

5. Originally Posted by Ep!taph

dy/dx + 3y = 6
By the method of separating the variables determine the equation that describes 'y' given that y = 0 when x =0
Note that $\displaystyle \frac{\,dy}{\,dx}+3y=6\implies\frac{\,dy}{\,dx}=6-3y$

Now separate the variables:

$\displaystyle \frac{\,dy}{3\left(2-y\right)}=\,dx$

Then integrate and solve for y:

$\displaystyle \int\frac{\,dy}{3\left(2-y\right)}=\int\,dx\implies-\tfrac{1}{3}\ln\left|2-y\right|=x+C$ $\displaystyle \implies\left|2-y\right|=Ce^{-3x}\implies y=2-Ke^{-3x}$

Now apply the initial condition $\displaystyle y\left(0\right)=0$ to find K.

Can you take it from here?

P.S.: From now on, please post a new question in a new thread.