Question

Find a particular solution to a given third-order homogeneous linear equation.

I am doing this as a bit of self study and really dont see how I am to solve this. if someone would be able to give me full reasoning behind the answer that would be great. Question below.

Answer 1

\[y^{(3)}+2y''-y'-2y=0; y(0)=1, y'(0)=2, y''(0)=0; y_1=e^x,y_2=e^{-x}, y_3=e^{-2x}\]

Answer 2

simplest way i think is to let D be the differential operator.

this is equal to \[D^3 + 2D^2 - D - 2 = 0\]

find the roots (used wolfram). D = -1, 1, 2

solutions: \[c_{1} e^{(-x) } + c_{2}e^{x} + c_{3}e^(2x)\]

Answer 3

The answer in the book is showing that it is \[y(x)=\frac{4}{3}e^x-\frac{1}{3}e^{-2x}\]
So by simply finding the roots how I can come up with 4/3, 0, -1/3; they do not seem to have any realtion to -1,1,2 that you posted.

Answer 4

I guess what you mean with a particular solution is assigning a value to the various constants that emerge from solving for a general solution for the given Differential Equation, however, if that is the case you only want to apply the initial conditions and solve the given 3x3 system for c_1, c_2, c_3

Answer 5

For instance y(0)=C_1 +C_2 + C_3 =0

Answer 6

Thanks guys, got it. :)

Answer 7

my bad. my answer is the general solution that you already had lol (y1, y2 and y3)