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: 

Now we apply basic factorization: 

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

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

where: 

 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

 , e^0 = 1

 (equation 1)

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

 (we know from above e^0 =1)

  (equation 2)

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

Now we can find C_2 from equation 1: 

Now we have our constants our particular equation is: