# Differential Equations...

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• Feb 12th 2009, 02:20 AM
mr_motivator
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?
• Feb 12th 2009, 02:49 AM
mr fantastic
Quote:

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.
• Feb 12th 2009, 10:04 AM
ThePerfectHacker
Quote:

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

Multiply by $e^{3x}$ to get $e^{3x}y' + 3e^{3x}y = 0 \implies \left( e^{3x}y \right)' = 0$
Therefore, $e^{3x}y = k \implies y = ke^{-3x},k\in \mathbb{R}$.
• Feb 12th 2009, 12:30 PM
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.
• Feb 12th 2009, 09:12 PM
ThePerfectHacker
Quote:

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 $y=-2$.