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Chapter 8: Problem 2

What is the general solution of the equation \(y^{\prime}(t)=3 y-12 ?\)

Short Answer

Expert verified
Question: Determine the general solution of the first-order linear differential equation \(y'(t) = 3y - 12\). Answer: The general solution of the given differential equation is \(y(t) = -4 + Ce^{3t}\), where \(C\) is an arbitrary constant.

Step by step solution

01

Rewrite the equation in standard form

To solve the given differential equation, we first rewrite it as a first-order linear differential equation. In standard form, we have: $$y'(t) - 3y = -12$$ This is now a first-order linear differential equation with \(P(t) = -3\) and \(Q(t) = -12\).
02

Find the integrating factor

To find the general solution to this equation, we will use an integrating factor. The integrating factor is given by \(e^{\int P(t) dt}\). In this case, our integrating factor is: $$e^{\int -3 dt} = e^{-3t}$$
03

Multiply the differential equation by the integrating factor

Now we multiply the given differential equation by the integrating factor: $$e^{-3t}(y'(t) - 3y) = -12e^{-3t}$$ This simplifies the left side of the equation to the derivative of \((ye^{-3t})\) with respect to \(t\): $$(ye^{-3t})' = -12e^{-3t}$$
04

Integrate both sides of the equation

Now we can integrate both sides of the equation to solve for \(y\): $$\int (ye^{-3t})' dt = \int -12e^{-3t} dt $$ On the left, the integral of the derivative is simply the function itself: $$ye^{-3t} = \frac{-4e^{-3t}}{1} + C$$ Now, to find the general solution, we can multiply both sides by \(e^{3t}\) to isolate \(y\):
05

Isolate y and write the general solution

Multiply both sides by \(e^{3t}\): $$y(t) = \frac{-4e^{-3t}}{1}e^{3t} + Ce^{3t}$$ This simplifies to: $$y(t) = -4 + Ce^{3t}$$ This is the general solution of the given differential equation. Here, \(C\) is an arbitrary constant which can be found if an initial condition is given.

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Most popular questions from this chapter

Solve the following initial value problems. When possible, give the solution as an explicit function of \(t\) $$y^{\prime}(t)=\frac{3 y(y+1)}{t}, y(1)=1$$

Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem $$P^{\prime}(t)=r P\left(1-\frac{P}{K}\right), \quad P(0)=P_{0}$$ $$\text { is } P(t)=\frac{K}{\left(\frac{K}{P_{0}}-1\right) e^{-n}+1}$$

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem and graph the solution to be sure that \(m(0)\) and \(\lim _{t \rightarrow \infty} m(t)\) are correct. A one-million-liter pond is contaminated and has a concentration of \(20 \mathrm{g} / \mathrm{L}\) of a chemical pollutant. The source of the pollutant is removed and pure water is allowed to flow into the pond at a rate of \(1200 \mathrm{L} / \mathrm{hr} .\) Assuming that the pond is thoroughly mixed and drained at a rate of \(1200 \mathrm{L} / \mathrm{hr}\), how long does it take to reduce the concentration of the solution in the pond to \(10 \%\) of the initial value?

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$\begin{array}{l} e^{-y / 2} y^{\prime}(x)=4 x \sin x^{2}-x ; y(0)=0 \\ y(0)=\ln \left(\frac{1}{4}\right), y(\sqrt{\frac{\pi}{2}})=0 \end{array}$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the growth rate function for a population model is positive, then the population is increasing. b. The solution of a stirred tank initial value problem always approaches a constant as \(t \rightarrow \infty\) c. In the predator-prey models discussed in this section, if the initial predator population is zero and the initial prey population is positive, then the prey population increases without bound.
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