Solve the following initial-value problem: y"+6y'+5y=0; y(0)=0. y'(0)=3

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To solve the given initial-value problem, we need to find the general solution for the differential equation y"+6y'+5y=0 and then apply the initial conditions y(0)=0 and y'(0)=3 to determine the specific solution.

Step 1: Finding the general solution
The differential equation is a second-order linear homogeneous equation, so we assume a solution of the form y = e^(rt), where r is a constant to be determined.

Plugging this assumption into the differential equation, we get:
r^2e^(rt) + 6re^(rt) + 5e^(rt) = 0

Factoring out e^(rt), we get:
e^(rt)(r^2 + 6r + 5) = 0

For the equation to hold true, either e^(rt) should be equal to zero (which is impossible) or (r^2 + 6r + 5) should be equal to zero.

Now we solve the quadratic equation r^2 + 6r + 5 = 0 for r:
(r+5)(r+1) = 0

From this equation, we obtain two distinct roots: r = -5 and r = -1.

So, the general solution of the given differential equation is:
y(t) = C1e^(-5t) + C2e^(-t)

Step 2: Applying the initial conditions
Now, we substitute the values of y(0) = 0 and y'(0) = 3 into the general solution to determine the specific solution.

Using y(0) = 0:
y(0) = C1e^(-5*0) + C2e^(-1*0)
0 = C1 + C2

Using y'(0) = 3:
y'(0) = -5C1e^(-5*0) - C2e^(-1*0)
3 = -5C1 - C2

We now have a system of two equations with two unknowns:
0 = C1 + C2
3 = -5C1 - C2

We can solve this system of equations to find the values of C1 and C2.

Multiplying the first equation by -1 and adding the two equations, we get:
3 = -4C1
C1 = -3/4

Substituting C1 = -3/4 into the first equation, we find:
-3/4 + C2 = 0
C2 = 3/4

So, the specific solution for the given initial-value problem is:
y(t) = (-3/4)e^(-5t) + (3/4)e^(-t)

Therefore, the solution to the initial-value problem y"+6y'+5y=0; y(0)=0, y'(0)=3 is y(t) = (-3/4)e^(-5t) + (3/4)e^(-t).