23 April, 20:00
Solve each of the following initial value problems and plot the solutions for several values of y0. Then describe in a few words how the solutions resemble, and differ from, each other.
(a) dy/dt = - y + 5, y (0) = y0
(b) dy/dt = - 2y + 5, y (0) = y0
(c) dy/dt = - 2y + 10, y (0) = y0
23 April, 20:54
a) y-8 = (y₀-8), b) 2y - 5 = (2y₀-5)
To solve these equations the method of direct integration is the easiest.
a) the given equation is
dy / dt = and - 8
dy / y-8 = dt
We change variables
y-8 = u
dy = du
We replace and integrate
∫ du / u = ∫ dt
Ln (y-8) = t
We evaluate at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Let's simplify the equation
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) the equation is
dy / dt = 2y - 5
u = 2y - 5
du = 2 dy
du / 2u = dt
½ Ln (2y-5) = t
We evaluate at the limits
½ [ln (2y-5) - ln (2y₀-5) ] = t
Ln (2y-5 / 2y₀-5) = 2t
2y - 5 = (2y₀-5)
c) the equation is very similar to the previous one
u = 2y - 10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
ln (2y-10) - ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
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» Solve each of the following initial value problems and plot the solutions for several values of y0. Then describe in a few words how the solutions resemble, and differ from, each other.