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

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`