Please check my work on initial value problem. thank you

[darryw]

Homework Statement

Solve the following initial value problem. Sketch the solution and describe its behavior
as t increases.
y'' + 4y' + 3y = 0

y(0) = 2
y'(0) = -1

1st i solved the characteristic:

r^2 + 4r + 3 = 0

r = -1
r = -3

then the general solution is;

y = c_1e^-x + c_2e^-3x
also..
y' = -c_1e^-x - 3c_2e^-3x

so after i plug in the initial values i get 2 equations and 2 unknowns..

c_1 + c_2 = 2
-c_1 - 3c_2 = -1

c_1 = 5/2
c_2 = -1/2

so the solution with initial conditions (is this called "particular solution?) is:

y = (5/2)e^-x - (1/2)e^-3x

is this much correct before i describe the equations behavior? thanks a lot

Homework Equations

The Attempt at a Solution

 

[Mark44]

Homework Statement

Solve the following initial value problem. Sketch the solution and describe its behavior
as t increases.
y'' + 4y' + 3y = 0

y(0) = 2
y'(0) = -1

1st i solved the characteristic:

r^2 + 4r + 3 = 0

r = -1
r = -3

then the general solution is;

y = c_1e^-x + c_2e^-3x
also..
y' = -c_1e^-x - 3c_2e^-3x

so after i plug in the initial values i get 2 equations and 2 unknowns..

c_1 + c_2 = 2
-c_1 - 3c_2 = -1

c_1 = 5/2
c_2 = -1/2

so the solution with initial conditions (is this called "particular solution?) is:

y = (5/2)e^-x - (1/2)e^-3x

is this much correct before i describe the equations behavior? thanks a lot

You've done all the hard work. Checking the solution to an initial value problem is easy in comparison. For your solution, y = (5/2)e^-x - (1/2)e^-3x, is it true that y(0) = 2 and y'(0) = -1? If so, then your solution satisifies the initial conditions, mean that the solution function goes through (0, 2) and its slope there is -1.

For your solution function, is it true that y'' + 4y' + 3y = 0? If so, then your solution satisifies the differential equation, and you're home free.