# Solve y''+4y'+3y=0, y(0)=2, y'(0)=-1

## Expert Answers We begin the question by finding the general solution of this second order differential equation. In order to approach this question we need to find the characteristic equation first:

The characteristic equation is found as follows:

`D^2 + 4D + 3 = 0`

Now we apply basic factorization:

`(D+3) (D+1) = 0`

Now we determine the roots by equating each term to zero:

`D+ 3 = 0 or D+1= 0`

`D = -3 or D = -1`

From the above roots we can now find the general solution:

`y (x) = C_1 e^(-3x) + C_2 e^(-x)`

where:

`C_1 and C_2` are constants.

Since we have conditions, y(0) = 2 and y'(0) = 1, we can find the particular solution and solve for the above constants.

Let's begin with the first constraint: y(0) = 2

`y(0) = C_1 e^(-3*0) +C_2 e^(-1*0)` , e^0 = 1

`2 = C_1 + C_2` (equation 1)

Now we use the second constraint y'(0)=1. But first we must find y'(x)

`y' (x) = -3C_1 e^(-3x) - C_2 e^-x`

`y'(0) = -3 C_1 -C_2` (we know from above e^0 =1)

`-1 = -3C_1 -C_2`  (equation 2)

We have two unknowns and two equations. We can now add both equations:

`2-1 = -3C_1 + C_1 -C_2 +C_2`

`1 = -2 C_1`

`C_1 = -1/2`

Now we can find C_2 from equation 1:

`C_2 = 2 - 1/2 = 5/2`

Now we have our constants our particular equation is:

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

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