Question

2dx/dt+dy/dt=5e^t & dy/dt-3dx/dt=5. y=0,x=0,t=0

Answer 1

2\(\frac{dx}{dt}\) + \(\frac{dy}{dt}\) = 5e^t .....(1)

\(\frac{dy}{dt}\) - 3\(\frac{dx}{dt}\) = 5 .....(2)

By subtracting equation (2) from (1), we get

5\(\frac{dx}{dt}\) = (e^t - 1)

\(\frac{dx}{dt}\) = e^t-1

dx = (e^t - 1)dt

x = e^t- t + c (By integrating both sides)

at t = 0, x = 0

∴ 0 = -e⁰ - 0 + c

c = -e⁰ = -1

∴ x = e^t - t -1

By putting x = e^t - t -1 in equation (2), we get

\(\frac{dy}{dt}\) = 5 + 3\(\frac{d}{dt}\) (e^t- t -1)

= 5 + 3e^t - 3

= 2 + 3e^t

dy = (2+3e^t)dt

y = 2t + 3e^t + c (By integrating both sides)given that at t = 0, y = 0

∴ 0 = 2 x 0 + 3e⁰ + c

c = -3e⁰ = -3

∴y = 2t + 3e^t - 3

Hence, solution of given system of differential equation is x = e^t - t - 1 and y = 2t + 3e^t - 3