Solving Second Order Differential Equations with Constant Coefficients

[thomas49th]

Homework Statement

I've completed this question but I don't 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 don't 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:

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

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 :)

 

[tiny-tim]

First of all i created auxilary equation

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

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

erm … -3 and -1

 

[thomas49th]

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

(T-2)²
when it should be +

so is the final answer:

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

 

[tiny-tim]

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

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

No! Try again.

so the general solution is:

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

is that right :) ?!

Cheers :)

No, the particular solution is added to the general solution.

 

[thomas49th]

so where does the complimentry solution come in?

Thanks :)

 

[tiny-tim]

so where does the complimentry solution come in?

Thanks :)

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

… the particular solution is added to the complementary solution.

 

[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 :)

 

[tiny-tim]

How do I find the values of A and B

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

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

You don't need to …

the general solution does have two unknown constants.

 

[thomas49th]

so is my final answer:

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

 

[tiny-tim]

so is my final answer:

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

No …

see your post #3 …

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! )

 

[Mark44]

so is my final answer:

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

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.)

[Answer Removed]

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.

 

[Mark44]

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

 

[tiny-tim]

No, your values for a and b, the coefficient of t and the constant, are wrong. …. 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.

Hey Mark …

no complete solutions, please!

could you edit? ​