# Introductuin to differential equations

1. Oct 18, 2007

### sapiental

1. The problem statement, all variables and given/known data
use the concept that y = c, -inf < x < inf is a constant function if and only if y' = 0 to determine whether the specified differential equation has any constant solutions:

y'' + 4y' + 6y = 10

3. The attempt at a solution

What throws me off in this particular problem is y''. If y' = 0 then y'' = 0 as well.
Only now when I integrate y'' to y' its y' = c1 so c1 = 0 and y' still = 0.

then I basically substitute my results in the equation to get:

(0) + 4(0) + 6(c) = 10

6c = 10

[c = 10/6, c1 = 0] are the constant solutions to the equation.

Does this make sense?

Thanks.

2. Oct 18, 2007

### Dick

You are thinking too hard. You don't have to integrate anything. Set y''=y'=0 and solve the equation for y. y=0 is NOT a solution.

3. Oct 18, 2007

### sapiental

ahh.. so

y = (10 - 4y' - y'')/6

y = (10 - 4(0) - 0)/6

[y = 10/6] ?

Thanks for the help! Please let me know if this is right.

4. Oct 18, 2007

### Kreizhn

The great thing about ODE's is that you can usually quite readily check the answer by substituting your solution back into the differential equation. Give it a try and you'll find that your solution is correct.