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