Question

Solve the differential equation ((d^3)y)/d(t^3)+ 2(((d^2)y)/(d(t^2))−(dt/dt)−2y= 0 where y= 1,dy/dt=2, d^2y/dt^2= 2 at t= 0.

Answer 1

D³y + 2D²y - Dy - 2y = 0, where D = \(\frac{d}{dt}\)

It's auxiliary equation is

m³ + 2m² - m - 2 = 0

⇒ m²(m + 2) - 1(m + 2) = 0

⇒ (m + 2) (m² - 1) = 0

⇒ m = -2, 1, 1

\(\therefore\) C.F. = C₁e^-2t + (C₂ + C₃t)e^t

\(\therefore\) y = C₁ e^-2t + (C₂ + C₃t)----(1)

\(\frac{dy}{dt}\) = -2C₁e^-2t + (C₂ + C₃ + C₃t)e^t---(2)

\(\frac{d^2y}{dt^2}\) = 4C₁e^-2t + (C₂ + 2C₃ + C₃e^t)---(3)

At t = 0, y = 1, \(\frac{dy}{dt}=2,\frac{d^2y}{dt^2}=2\)

Put t = 0 in equations(1), (2) and (3) we get

C₁ + C₂ = 1---(4)

-2C₁ + C₂ + C₃ = 5 ---(5)

4C₁+ C₂ + 2C₃ = 2 ---(6)

substract equation(4) from (5) and (6) we get

-3C₁ + C₃ = 1---(7)

3C₁ + 2C₃ = 1--(8)

On adding equations (7) and (8), we get

3C₃ = 2

⇒ C₃ = 2/3

Then from (7) we get

-3C₁ + 2/3 = 1

⇒ -3C₁  = 1 - 2/3 = 1/3

⇒ C₁ = -1/9

Then from (4) we get

C₂ = 1 - C₁ = 1 + 1/9 = 10/9

\(\therefore\) y = -1/9 e^-2f + (10/9 + 2/3 t)e^t

 which is solution of given differential equation.