1. ## Differential Equations...

a) Write down the general solution of the differential equation dy/dx + 3y = 0.

b) Use a suitable trial function to find one simple particular solution of dy/dx + 3y = -6

c) Find the solution of dy/dx + 3y = -6 that satisfies y(0) = -5

Any ideas?

2. Originally Posted by mr_motivator
a) Write down the general solution of the differential equation dy/dx + 3y = 0.

b) Use a suitable trial function to find one simple particular solution of dy/dx + 3y = -6

c) Find the solution of dy/dx + 3y = -6 that satisfies y(0) = -5

Any ideas?
a) This is first order linear. What do your notes say to do?

b) Try a solution of the form y = k. What value do you find k has to have?

c) y = Solution to a) + solution to b). Now substitute x = 0 and y = -5 to solve for the arbitrary constant.

3. Originally Posted by mr_motivator
a) Write down the general solution of the differential equation dy/dx + 3y = 0.
Multiply by $\displaystyle e^{3x}$ to get $\displaystyle e^{3x}y' + 3e^{3x}y = 0 \implies \left( e^{3x}y \right)' = 0$
Therefore, $\displaystyle e^{3x}y = k \implies y = ke^{-3x},k\in \mathbb{R}$.

4. so the answer to a), the general solution to the differential equation is y(x) = kexp(-3x) ?

Or is this the answer to b), the trial function? If anyone knows a good tutorial for the use of trial functions in this context it would be appreciated as i cannot seem to find any.

5. Originally Posted by Muzza8888
so the answer to a), the general solution to the differential equation is y(x) = kexp(-3x) ?

Or is this the answer to b), the trial function? If anyone knows a good tutorial for the use of trial functions in this context it would be appreciated as i cannot seem to find any.
For (b) just take $\displaystyle y=-2$.