To find the general solution of the second-order differential equation:
y” + 8y’ + 17y = 0
we first rewrite it in standard form:
1. **Characteristic Equation**: We start by forming the characteristic equation associated with the differential equation. This is done by substituting y = e^{rt}, where r is a constant. We differentiate this substitution to find:
– The first derivative: y’ = re^{rt}
– The second derivative: y” = r^2e^{rt}
The left-hand side of the differential equation then becomes:
r^2e^{rt} + 8re^{rt} + 17e^{rt} = 0
Factoring out e^{rt}, we have:
e^{rt}(r^2 + 8r + 17) = 0
Since e^{rt} is never zero, we set the characteristic polynomial equal to zero:
r^2 + 8r + 17 = 0
2. **Finding Roots**: To solve for r, we use the quadratic formula:
r = rac{-b ext{±} ext{sqr}(b^2 – 4ac)}{2a}
where a = 1, b = 8, and c = 17.
Calculating the discriminant:
b^2 – 4ac = 8^2 – 4(1)(17) = 64 – 68 = -4
Since the discriminant is negative, we conclude that the roots are complex:
r = rac{-8 ext{±} ext{i}2}{2} = -4 ext{±} i
3. **General Solution**: The general solution for a second-order linear differential equation with complex roots r = eta ext{±} i heta is given by:
y(t) = e^{eta t}(C_1 ext{cos}( heta t) + C_2 ext{sin}( heta t))
where C_1 and C_2 are constants determined by initial conditions. In our case:
Substituting eta = -4 and heta = 2 into the formula gives:
y(t) = e^{-4t}(C_1 ext{cos}(2t) + C_2 ext{sin}(2t))
The general solution of the given differential equation is thus:
y(t) = e^{-4t}(C_1 ext{cos}(2t) + C_2 ext{sin}(2t))