# Second order differential

1. Nov 28, 2008

### thomas49th

1. The problem statement, all variables and given/known data
I've completed this question but I dont think i've got it right. It's from a 2008 June FP1 further maths paper. There is a second part to the question I dont understand which I will post after I know I have got this part right :)

Find, in terms of k, the general solution of the differential equation:

$$\frac{d^{2}x}{dt} + 4 \frac{dx}{dt} + 3x = kt + 5$$

3. The attempt at a solution

First of all i created auxilary equation

$$T^{2} + 4T + 3$$

and solved it to get 2 real roots, 3 and 1

2 real roots imply that the complementory soltuion:

is $$y = Ae^{çx} + Be^{ßx}$$
i can fill in ç and ß, as it's the roots from the auxilary equation
$$y = Ae^{x} + Be^{3x}$$

Now I do need use inspection to find the particular soltution:

$$\frac{d^{2}x}{dt} + 4 \frac{dx}{dt} + 3x = kt + 5$$

let x = at + b

$$\frac{d^{2}x}{dt} = 0 \frac{dx}{dt} = a$$

try:

$$0 + 4(a) + 3(at +b) \eqiv kt + 5$$

compare coefficients

so 3a = k
4a + 3b = 5

a = k/3
b = (15-4k)/3

so the general solution is:

$$y = \frac{k}{3} e^{x} + \frac{15-4k}{3} e^{3x}$$

is that right :) ?!

Cheers :)

2. Nov 28, 2008

### tiny-tim

erm … -3 and -1

3. Nov 28, 2008

### thomas49th

Sorry im being an idiot. I used comleting the square and by habbity i did

(T-2)²
when it should be +

$$y = \frac{k}{3} e^{-x} + \frac{15-4k}{3} e^{-3x}$$

4. Nov 28, 2008

### tiny-tim

No! Try again.
No, the particular solution is added to the general solution.

5. Nov 28, 2008

### thomas49th

so where does the complimentry solution come in?

Thanks :)

6. Nov 28, 2008

### tiny-tim

oh … i meant complementary … i couldn't remember the word

… the particular solution is added to the complementary solution.

7. Nov 28, 2008

### thomas49th

aha, cool.
so
$$y = Ae^{x} + Be^{3x}$$
+

a = k/3
b = (15-4k)/3

is that it? How do I find the values of A and B

Thanks :)

8. Nov 28, 2008

### tiny-tim

You don't need to

the general solution does have two unknown constants.

9. Nov 30, 2008

### thomas49th

$$y = Ae^{x} + Be^{3x} + \frac{k}{3} + \frac{15-4k}{3}$$

10. Nov 30, 2008

### tiny-tim

No …

and i don't think the fractions are right either …

please get into the habit of plugging your solution back into the original equation, to see if it works!

(if it doesn't, then its wrong! )

11. Nov 30, 2008

### Staff: Mentor

No, your values for a and b, the coefficient of t and the constant, are wrong. They should be: a = k/3, b = 5/3 - 4k/9. And you forgot that the coefficients of x in the exponential terms are negative. (I have reverted to the variables of the original problem, x and t.)

The homogeneous solution (aka complementary solution), $$x_h(t)$$ is the part of the general solution for which x''(t) + 4x'(t) + 3x(t) = 0. The particular solution is the part of the general solution for which x''(t) + 4x'(t) + 3x(t) = kt + 5.

One thing about differential equations: getting the solutions takes a bit of doing, but if you know how to differentiate, you can check that your answer actually is a solution by verifying that the appropriate combination of x''(t), x'(t), and x(t) actually add up to what they're supposed to.

Last edited by a moderator: Nov 30, 2008
12. Nov 30, 2008

### Staff: Mentor

BTW, for future reference, differential equations are NOT precalculus!

13. Nov 30, 2008

Hey Mark …