Question

Solve each of the following initial value problems and plot the solutions for several values of y0. (A computer algebra system is recommended.) (a) dy/dt = y−8, y(0) = y0(b) dy/dt = 2y−5, y(0) = y0(c) dy/dt = 2y−10, y(0) = y0

Answer 1

Answer:

a) y-8 = (y₀-8) $e^{t}$ , b) 2y -5 = (2y₀-5) $e^{2t}$

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 = $e^{t}$

             y-8 = (y₀-8) $e^{t}$

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) $e^{2t}$

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) $e^{2t}$

Answer 2

Integration is a method to add the parts to find the whole. The value of $y_{(0)}$

• A)$y-8=(y_0-8)e^t$

• B)$2y-5=(2y_0-5)e^{2t}$

• c)$2y-10=(2y_0-10)e^{2t}$

Given information-

The equation given in the problem is,

$\dfrac{dy}{dt} =y-8$

$\dfrac{dy}{dt} =2y-5$

$\dfrac{dy}{dt} =2y-10$

What is integration?

Integration is a method to add the parts to find the whole.

• A) Given equation in part A is,

$\dfrac{dy}{dt} =y-8\\\dfrac{dy}{y-8}=dt$                      ......1

Let,

$y-8=u\\dy=du$

Put the values in equation 1 and integrate both sides,,

$\int\dfrac{du}{u}=\int dt\\\ln u=t$

Put the values beck,

$ln(y-8)=t$

For the values of $y_0$, evaluate with lower limit as $t$ equal to zero,

$\ln\dfrac{(y-8)}{(y_0-8)} =t\\y-8=(y_0-8)e^t$

• B) Given equation in part B is,

$\dfrac{dy}{dt} =2y-5\\\dfrac{dy}{2y-5}=dt$                      ......1

Let,

$2y-5=u\\2dy=du$

Put the values in equation 1 and integrate both sides,,

$\int\dfrac{du}{2u}= \int dt\\\dfrac{1}{2}\ln u=t$

Put the values beck,

$\dfrac{1}{2} \ln(2y-5)=t$

For the values of $y_0$, evaluate with lower limit as $t$ equal to zero,

$\dfrac{1}{2} \ln\dfrac{(2y-5)}{(2y_0-5)} =t\\2y-5=(2y_0-5)e^{2t}$

• C) Given equation in part C is,

$\dfrac{dy}{dt} =2y-10\\\dfrac{dy}{2y-10}=dt$                      ......1

Let,

$2y-5=u\\2dy=du$

Put the values in equation 1 and integrate both sides,,

$\int\dfrac{du}{2u}= \int dt\\\dfrac{1}{2}\ln u=t$

Put the values beck,

$\dfrac{1}{2} \ln(2y-10)=t$

For the values of $y_0$, evaluate with lower limit as $t$ equal to zero,

$\dfrac{1}{2} \ln\dfrac{(2y-10)}{(2y_0-10)} =t\\2y-10=(2y_0-10)e^{2t}$

Hence the value of $y_{(0)}$

• A)$y-8=(y_0-8)e^t$

• B)$2y-5=(2y_0-5)e^{2t}$

• c)$2y-10=(2y_0-10)e^{2t}$

Learn more about the integration here;

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