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: 

0

Now we apply basic factorization: 

0

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

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

e^(-x)

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

= C_1 + C_2

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: 

C_1

Now we can find C_2 from equation
1: 

Now we have our
constants our particular equation is: 

   

e^(-3x) + (5/2)e^-x