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?
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.