To find the general solution of the second order differential equation y” + 8y’ + 17y = 0, we start by solving the characteristic equation associated with the differential equation.
The characteristic equation can be obtained by substituting y = e^{rt} into the differential equation, which leads to:
r2 + 8r + 17 = 0
Next, we can use the quadratic formula to find the roots of this equation:
r = rac{-b \u2013 000 000 ext{sqrt}(b^2 4ac)}{2a}
Here, a = 1, b = 8, and c = 17. Plugging in these values, we have:
r = rac{-8 60 ext{sqrt}(8^2 4(1)(17))}{2(1)} = rac{-8 60 ext{sqrt}(64 68)}{2} = rac{-8 60 ext{sqrt}(-4)}{2}
This simplifies to:
r = rac{-8 60 2i}{2} = -4 60 i
The roots are complex: r = -4 60 i. For complex roots of the form r = eta 60 i, the general solution of the differential equation is given by:
y(t) = C_1 e^{eta t} ext{cos}(
u t) + C_2 e^{eta t} ext{sin}(
u t)
In our case, since we have r = -4 60 i, it implies eta = 0 and
u = 4. Thus, the solution becomes:
y(t) = C_1 e^{0} ext{cos}(4t) + C_2 e^{0} ext{sin}(4t)
Which simplifies to:
y(t) = C_1 ext{cos}(4t) + C_2 ext{sin}(4t)
Here, C_1 and C_2 are arbitrary constants determined by initial conditions. This expression represents the general solution of the given second order differential equation.