Question

Solve the differential equations (dx/dt)+(dy/dt)=t and ((d^2)x)/(d(t^2))−y=e^t given x= 3, dx/dt=−2, y= 0 when t= 0.

Answer 1

\(\frac{dx}{dt}+\frac{dy}{dt} = t\)

⇒ Dx + Dy = t---(1), where d/df ≡ D

\(\frac{d^2x}{dt^2}-y=e^t\) 

⇒ D²x - y = e^t

⇒ D³x - Dy = e^t ---(2)

By adding (1) and (2), we get

D³x + Dx = e^t + t---(3)

It's auxiliary equation is

m³ + m = 0

⇒ m(m² + 1) = 0

⇒ M = 0, \(\pm i\)

\(\therefore\) C.F. = C₁ e^ot + C₂ cos t + C₃ sin t

 = C₁ + C₂ cos t + C₃ sin t

P.I. = \(\frac1{D^3+D}(e^t+t)\) 

 = \(\frac{e^t}{1^3+1}+\frac1{D}(1+D^2)^{-1}t\) 

 = \(\frac{e^t}2+\frac1{D}(1-D^2+D^4-....)t\) 

 = \(\frac12e^t+(\frac1D-D+D^3-....)t\)

= \(\frac12e^t+\frac{t^2}2-1\) 

\(\therefore\) x = C.F. + P. I.

 = C₁ + C₂ cos t + C₃ sin t + \(\frac12\)e^t + \(\frac {t^2}2-4\)---(4)

\(\therefore\) \(\frac{dx}{dt}\) = - C₂sin t + C₃ cos t + \(\frac12\)e^t + t---(5)

Then from (1) we get,

\(\frac{dy}{dt}\)  - C₂ sin t - C₃ cos t + \(\frac12\)e^t + t = t

⇒ \(\frac{dy}{dt}\) =  C₂ sin t - C₃ cos t - \(\frac12\)e^t

\(\therefore\) y = -C₂ cos t - C₃ sin t - \(-\frac12\)e^t---(6)

Given that when t = 0 we have y = 0, x = 3 and \(\frac{dx}{dt}=-2\) 

\(\therefore\) -C₂ - \(\frac12\) = 0 (From (6))

⇒ C₂ = -1/2

C₃ + 1/2 = -2 (From (5))

C₃  = -2 - 1/2 = -5/2

And C₁ + C₂ + 1/2 - 1 = 3 (From (4))

⇒ C₁ - 1/2 + 1/2 = 3 + 1

⇒ C₁ = 4

\(\therefore\) x = 4 - 1/2 cos t - 5/2 sin t + 1/2 e^t + t²/2 - 1

y = 1/2 cos t 5/2 sin t - 1/2 e^t

which is solution of given system of differential equations.