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### Messages - Ruoqi Deng

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##### Quiz-4 / Re: Q4 TUT 5102
« on: October 26, 2018, 08:51:27 PM »
y1(t)=e^t
y'1(t)=e^t
y''(t)=e^t

y2(t)=t
y'2(t)=1
y''2(t)=0

since (1-t)e^t * e^t + t * e^t - e^t =0 and (1-t)t + 1 - 0 = 0, y1(t) and y2(t) satisfy the corresponding homogeneous equation.

y'' + (t/1-t)y' - (1/1-t)y = -2(t-1)e^-t

g(t) = -2(t-1)e^-t

w[y1,y2] = y1(t)y2'(t) - y2(t)y1'(t) = (1- t)e^t

Y(t） = u1(t)y1(t) +u2(t)y2(t)

u1(t) = -$\int(y2(t)g(t))/W[y1,y2](t)])dt$= - $\int(t*(-2(t-1)e^-t)/(1-t)e^t)dt$ = -2$/int(te^(-2t))dt$ = (t + 1/2)e^(-2t)

u2(t) = $\int(y1(t)g(t))/W[y1,y2](t)])dt$ = -2e^-t

Y(t) =  (t + 1/2)e^(-2t) ]*e^t + (-2e^-t) * t = (1/2 - t)e^-t

The general solution is y(t) = c1e^t + c2t + (1/2 - t)e^-t

The particular solution is Y(t) = (1/2 - t)e^-t

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##### Quiz-4 / Re: Q4 TUT 5102
« on: October 26, 2018, 08:21:52 PM »
y1(t)=e^t
y'1(t)=e^t
y''(t)=e^t

y2(t)=t
y'2(t)=1
y''2(t)=0

since (1-t)e^t * e^t + t * e^t - e^t =0 and (1-t)t + 1 - 0 = 0, y1(t) and y2(t) satisfy the corresponding homogeneous equation.

y'' + (t/1-t)y' - (1/1-t)y = -2(t-1)e^-t

g(t) = -2(t-1)e^-t

w[y1,y2] = y1(t)y2'(t) - y2(t)y1'(t) = (1- t)e^t

Y(t） = u1(t)y1(t) +u2(t)y2(t)

u1(t) = -∫(y2(t)g(t))/W[y1,y2](t)])dt= - ∫(t∗(−2(t−1)e−t)/(1−t)et)dt = -2/int(te(−2t))dt = (t + 1/2)e^(-2t)

u2(t) = ∫(y1(t)g(t))/W[y1,y2](t)])dt = -2e^-t

Y(t) =  (t + 1/2)e^(-2t) ]*e^t + (-2e^-t) * t = (1/2 - t)e^-t

The general solution is y(t) = c1e^t + c2t + (1/2 - t)e^-t

The particular solution is Y(t) = (1/2 - t)e^-t

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