Q.426. Differential Equation

Question 426.
Find the particular solution of the differential equation,
dy/dt = 0.5-t + 2y,  y(0) = 2.

Answer 426.
dy/dt = 0.5 - t + 2y
=> dy/dt - 2y = 0.5 - t
=> e^-2t dy/dt - 2y e^-2t = (0.5 - t)e^(-2t)
=> d(ye^-2t) = (0.5 - t)e^(-2t) dt

Integrating,
y e^(-2t)
= ∫ (0.5 - t) e^(-2t) dt
= (0.5) ∫ e^(-2t) dt - ∫ t e^(-2t) dt
= - (1/4) e^(-2t) - [ t ∫ e^(-2t) dt - ∫ [d/dt(t) ∫ e^(-2t) dt] dt]
= - (1/4) e^(-2t) - [- (1/2) te^(-2t) + (1/2) ∫ e^(-2t) dt]
= - (1/4) e^(-2t) - [- (1/2) te^(-2t) - (1/4) e^(-2t)] + c/2
= (1/2) te^(-2t) + c/2
=> ye^(-2t) = (1/2) te^(-2t) + c/2
=> 2y = t + ce^(2t)

Plugging t = 0 and y = 2
=> 4 = c
=> particular solution is
2y = t + 4e^(2t)
=> y = t/2 + 2e^(2t).

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