Question

Solve the initial-value problem:
2y'' + 5y' + 3y = 0 , y (0) = 3, y'(0) = -4

Answer 1

The auxillary equation would then be:

2r^2 + 5r + 3 = 0

Using the quadratic formula:

r = -3/2 , r = -1

so
\[\large y = c_1 e^{-3/2~x} + c_2e^{-x}\]

What do i do next?...

Answer 2

we can do this either undetermined nevermind ... it's a constant coefficent problem

Answer 3

so ok first take the derivative of y that y you just found

Answer 4

okay.
so
\[y = -3/2c_1e^{-3/2x} - c_2e^{-x}\]
right?

Answer 5

y'(x)* i should put

Answer 6

yeah

Answer 7

ok so y(0)=3 is an initial value problem condition so we have it in the form of y(x_o)=yo so the x is equal to 0 and the y is equal to 3 for the ORIGINAL Y

Answer 8

Oh, so i just plug in!

Answer 9

yeah and then find c1 and c2.. afterwards plug in the value of c1 and c2 in the original equation

Answer 10

you can check this with laplace if you know how to do it... it would be easy because the laplace transformation of 0 is umm 0 :P

Answer 11

Not sure i would do it with laplace i'm just skipping around topics with calculus xD