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Answer: t = 10 min. (ii) Write the initial value problem for Q(t) (before the tank is empty) and solve it.
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(i) How long (in min) will it take for the tank to become empty?
(ii) Write the initial value problem for Q(t) (before the tank is empty) and solve it.
Q(t) =
(10 − t)^2 (t − cos(t) + 2).
(2xy + y^3 )dx + (x^2 + 3xy^2 − 2 y)dy = 0.
x^2 y + xy^3 − y^2 = C.
dx¯ dt
¯x.
x ¯ = c 1
cos(t) + sin(t) cos(t)
e^5 t^ + c 2
cos(t) − sin(t) − sin(t)
e^5 t.
(ii) Draw the phase portrait of the system. What is the origin called in this case? Is the origin stable, unstable or semi-stable?
y′′^ − 4 y′^ +13y = (4t−4) cos(3t)+(12t−6) sin(3t), y(0) = 1, y′(0) = 1.
y(t) = t cos(3t) + e^2 t^ cos(3t) −
e^2 t^ sin(3t).
y′′^ + 4y = g(t), y(0) = 0, y′(0) = 0,
where
g(t) =
1 , t < π, sin(t), t ≥ π.
The sine Fourier series of f (x) = 2x + 1 over [0, 1] is
n=
nπ
sin(2nπx) +
n=
[2n − 1]π
sin((2n − 1)πx).
You could have written the cosine Fourier series as well.
uxx = 4ut, 0 < x < π, t > 0;
u(0, t) = 10, ux(π, t) = 0, t > 0;
u(x, 0) = sin(3x/2) + 10, 0 < x < π.
Hint: Here the left end of the bar is kept at fixed temperature 10, while the right end is insulated. There is no ”ready formula” for this case, you have to work from scratch.
u(x, t) = e−^9 t/^16 sin(
3 x 2
uxx = utt, 0 < x < π, t > 0;
u(0, t) = 0, u(π, t) = 0, t ≥ 0;
u(x, 0) = sin(4x) + 2 sin(6x), ut(x, 0) = sin(10x), 0 ≤ x ≤ π.
u(x, t) = sin(4x) cos(4t) + 2 sin(6x) cos(6t) +
sin(10x) sin(10t).