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

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1. 23 April, 20:54
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Step-by-step explanation:

a) y-8 = (y₀-8), b) 2y - 5 = (2y₀-5)

Explanation:

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

We integrate

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

We evaluate

ln (2y-10) - ln (2y₀-10) = 2t

2y-10 = (2y₀-10)