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Chapter 8: Problem 15

Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots .\) to denote arbitrary constants. $$y^{\prime}(t)=3+e^{-2 t}$$

Short Answer

Expert verified
Question: Determine the general solution of the first-order differential equation: \(y^{\prime}(t)=3+e^{-2t}\) Answer: The general solution of the given differential equation is: \(y(t) = 3t -\frac{1}{2}e^{-2t} + C\)

Step by step solution

01

Identify the given differential equation

The given first-order differential equation is: $$y^{\prime}(t)=3+e^{-2 t}$$
02

Integrate both sides with respect to \(t\)

To find the general solution, we will integrate both sides: $$\int y^{\prime}(t) dt = \int (3 + e^{-2t}) dt$$
03

Perform integration

Integration of the left side, we have: $$y(t) = \int(3 + e^{-2t}) dt$$ Now, we can integrate the right side by splitting the integral: $$y(t) = \int 3 dt + \int e^{-2t} dt$$ Integrate 3 with respect to t: $$\int 3 dt = 3t + C_1$$ Integrate \(e^{-2t}\) with respect to t by using substitution method (let \(u= -2t\) and \(dt = -\frac{1}{2}du\)). $$\int e^{-2t} dt = -\frac{1}{2}\int e^u du=-\frac{1}{2}e^{u}=-\frac{1}{2}e^{-2t}+C_2$$
04

Combine the integrals and add the arbitrary constants

Add the integrals and constants together to find the general solution: $$y(t) = 3t + C_1 -\frac{1}{2}e^{-2t}+C_2$$ Combine arbitrary constants \(C_1\) and \(C_2\) into a single arbitrary constant \(C\): $$y(t) = 3t -\frac{1}{2}e^{-2t} + C$$
05

Write the final general solution

The general solution of the given differential equation is: $$y(t) = 3t -\frac{1}{2}e^{-2t} + C$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrals
The concept of integrals is fundamental in calculus and is the inverse process of differentiation. When we talk about integrating a function, we are looking to find a function whose derivative is the given function. This process is known as anti-differentiation.
In the exercise above, integration is used to solve the differential equation \(y'(t) = 3 + e^{-2t}\). We integrate both sides with respect to \(t\), which involves finding an antiderivative for each term on the right-hand side.
  • The integral of \(3\) with respect to \(t\) is \(3t + C_1\), where \(C_1\) is an arbitrary constant.
  • For \(e^{-2t}\), a substitution method is used. Substituting \(u = -2t\), the integral becomes \(-\frac{1}{2}e^{-2t} + C_2\), where \(C_2\) is another arbitrary constant.
Combining these integrals gives you the antiderivative needed to solve the differential equation, resulting in the general solution. Each integral reflects a summation aspect of calculus, indicating accumulated change or total quantity.
First-Order Differential Equations
First-order differential equations involve derivatives of the first degree鈥攖he simplest type of differential equations. These equations form the foundation upon which more complex equations are built.
In the given exercise, the differential equation \(y'(t) = 3 + e^{-2t}\) can be classified as a first-order differential equation because it includes the first derivative of \(y(t)\), that is, \(y'(t)\).
  • The form of a typical first-order differential equation is \( \frac{dy}{dt} = f(t, y) \). However, in this exercise, \(y'(t) = 3 + e^{-2t} \) is a simpler representation because \(f(t,y)\) depends only on \(t\) and not on \(y\).
  • Solving such equations often requires integration, as in this example, where integrating both sides with respect to \(t\) helps isolate \(y(t)\).
Such equations are crucial for modeling natural phenomena, where a rate of change in one variable is proportional to or influenced by another.
General Solution
The concept of a general solution is essential in the study of differential equations. It represents not just a single answer, but a family of solutions comprising all possible particular solutions. Each particular solution is obtained by assigning specific values to the arbitrary constants.
In our exercise, after integrating to solve the differential equation \(y'(t) = 3 + e^{-2t}\), we derive the general solution \(y(t) = 3t - \frac{1}{2}e^{-2t} + C\). Here, \(C\) is an arbitrary constant encompassing the constants \(C_1\) and \(C_2\) initially introduced during integration.
  • The general solution highlights the infinite possibilities of particular solutions that depend on different initial conditions or additional information.
  • Representing solutions in this flexible format aids in understanding how a system may behave under various circumstances.
The general solution emphasizes the broad applicability and adaptability of mathematical models in different contexts and scenarios.

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Most popular questions from this chapter

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. A pot of boiling soup \(\left(100^{\circ} \mathrm{C}\right)\) is put in a cellar with a temperature of \(10^{\circ} \mathrm{C}\). After 30 minutes, the soup has cooled to \(80^{\circ} \mathrm{C}\). When will the temperature of the soup reach \(30^{\circ} \mathrm{C} ?\)

Determine whether the following equations are separable. If so, solve the initial value problem. $$\frac{d y}{d x}=e^{x-y}, y(0)=\ln 3$$

The Gompertz growth equation is often used to model the growth of tumors. Let \(M(t)\) be the mass of a tumor at time \(t \geq 0 .\) The relevant initial value problem is $$ \frac{d M}{d t}=-r M \ln \left(\frac{M}{K}\right), M(0)=M_{0} $$ a. Graph the growth rate function \(R(M)=-r M \ln \left(\frac{M}{K}\right)\) (which equals \(M^{\prime}(t)\) ) assuming \(r=1\) and \(K=4 .\) For what values of \(M\) is the growth rate positive? For what value of \(M\) is the growth rate a maximum? b. Solve the initial value problem and graph the solution for \(r=1, K=4,\) and \(M_{0}=1 .\) Describe the growth pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor? c. In the general solution, what is the meaning of \(K ?\) where \(r\) and \(K\) are positive constants and \(0 < M_{0} < K\)

Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=y\left(4 t^{3}+1\right), y(0)=4$$

Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\) a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that $$ \frac{d y}{d x}=\frac{-2 x}{y} $$ b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{C}|x|\) and then explain why it follows that \(y^{2}=k x\) Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\)
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