Question

How would you do the following initial value problem d^2y/dt^2 + dy/dt = 2t where t is greater then 0 and y(0) = 0 and dy/dx (0) = 1

Answer 1

Bring the 2t across onto the same side, and calculate the eigen values, have you tried this?

Answer 2

what do you mean by calculate the eigen values??

Answer 3

are you comfy with me using y' to represent dy/dt? and y''=d^2y/dt^2?

Answer 4

etc

Answer 5

ya

Answer 6

alright when you have y^(n)+y^(n-1)+...+y'+y=q, where q is a function interms of t, then you can rewrite this as x^n+x^(n-1)+...+x+1=0, find the roots of x and you will find the eigen values, this corresponds to a complimentary solution, to get the general solution and thus solve the initial value problem you need to combine it with one particular solution

Answer 7

i.e.x^2+x=0 thus x(x+1)=0 and this corresponds to y[cf]= A+Be^(-t), the particular solution is obtained by substituting this into the equation and solving.

Answer 8

is this clear?

Answer 9

ya thanks

Answer 10

Eigenvalues are usually referred to in the context of systems of ODEs. You may be more familiar with "roots to the characteristic equation."

Alternatively you can solve via the Laplace transform, if you're familiar with that.