Solve the subsequent IVP.

y'' - 4y' + 9y = 0, y(0) = 0, y'(0) = -8

Solution

The characteristic equation for such differential equation is. As:

 r² - 4r + 9 = 0

 The roots of this equation are r_1,2  = 2 + √(5i). So the general solution to the differential equation is as:

y(t) = c₁ e^2t cos (√5t)+ c₂ e^2t sin (√5t)

Here, you'll note that we didn't differentiate it right away as we did in the previous section. The motive for this is easy. But the differentiation is not terribly complicated this can find a little messy. Thus, first looking at the initial conditions we can notice from the first one which if we just applied it we would find the subsequent.

0 = y (0) + c₁

Conversely, the first term will drop out so as to meet the first condition. It makes the solution, with its derivative as

y(t) = c₂ e^2t sin (√5t)

y'(t) = 2c₂ e^2t sin (√5t) +√5 c₂ e^2t cos (√5t)

A much fine derivative than if we'd complete the original solution. Here, apply the second initial condition to the derivative to find out,

-8 = y'(0) = √5 c₂                   ⇒ c₂ = -8/√5

The actual solution is here as:

y(t) =  -8/√5 e^2t sin (√5t)