1. ## differential equations

Can anyone help with this?

x''+9x'+20x=29+20t+6e^(-2t)

i) Obtain the solution x(t) as the sum of the solutions obtained when each component is considered individually in turn.
ii) Solve the equation directly

2. Originally Posted by boab
Can anyone help with this?

x''+9x'+20x=29+20t+6e^(-2t)

i) Obtain the solution x(t) as the sum of the solutions obtained when each component is considered individually in turn.
they are asking you to solve

$x'' + 9x\ + 20x = 29$

$x'' + 9x\ + 20x = 20t$

and

$x'' + 9x\ + 20x = 6e^{-2t}$

separately, then add the solutions together

the method of undetermined coefficients is the way to go here

ii) Solve the equation directly
see here

3. Thanks for your help but it hasn't really cleared anything up for me. I'm still not sure how to go about it. Can you offer any further help?

Oh, and thanks again for your help with another problem I posted

4. Originally Posted by boab
Thanks for your help but it hasn't really cleared anything up for me. I'm still not sure how to go about it. Can you offer any further help?

Oh, and thanks again for your help with another problem I posted
take $x'' + 9x' + 20x = 6e^{-2t}$ for example.

we must find the homogeneous solutions first. that is, solve $x'' + 9x' + 20x = 0$

we assume a solution of $x = e^{mt}$

the characteristic equation is $m^2 + 9m + 20 = 0$ .........you should know how to get this.

$\Rightarrow (m + 5)(m + 4) = 0$

$\Rightarrow m = -5$ or $m = -4$

Thus, the homogeneous solutions are: $x = Ae^{-5t} + Be^{-4t}$.

Now we find a particular solution. It must be of the form of the original right hand side, but not the same as the solution to the homogeneous. there is no danger of the latter case here.

so, assume $x_p = Ce^{-2t}$ is a particular solution.

$\Rightarrow x_p' = -2Ce^{-2t}$

and $x_p'' = 4Ce^{-2t}$

plug these into the original differential equation, we get:

$4Ce^{-2t} - 18Ce^{-2t} + 20Ce^{-2t} = 6e^{-2t}$

now simplify and equate coefficients to solve for C.

the solution to the equation will be of the form:

$x(t) = Ae^{-5t} + Be^{-4t} + x_p$

the others are solved in exactly the same way

5. Thanks again. I'll freely admit this is a part of the course which just isn't clicking (as you've probably gathered!). I'll see how I get on, but thanks again for your help.