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Chapter 9: Problem 4

Explain how to solve a separable differential equation of the form \(g(t) y^{\prime}(t)=h(t)\).

Short Answer

Expert verified
Question: Solve the separable differential equation \(g(t)y'(t) = h(t)\) and obtain its general solution. Answer: The general solution for the given separable differential equation is \(y(t) = K \cdot e^{\int \frac{h(t)}{g(t)} dt}\), where \(K\) is an arbitrary constant.

Step by step solution

01

Rewrite the equation to separate the variables

Divide both sides of the equation by \(g(t)\), and we get: \[y'(t) = \frac{h(t)}{g(t)}\] Now, we want to have all the terms involving \(y(t)\) on one side of the equation and all the terms involving \(t\) on the other side. To do this, we can rewrite the equation as: \[\frac{dy(t)}{dt} = \frac{h(t)}{g(t)}\]
02

Divide both sides by \(y\)

To separate the variables, we want to have \(y\) and \(t\) on opposite sides of the equation. We can achieve this by dividing both sides by \(y\): \[\frac{1}{y(t)}\frac{dy(t)}{dt} = \frac{h(t)}{g(t)}\]
03

Integrate both sides

Now that we have separated the variables, we can integrate both sides of the equation with respect to \(t\). We get: \[\int \frac{1}{y(t)}\frac{dy(t)}{dt} dt = \int \frac{h(t)}{g(t)} dt\] Using the substitution method, we can simplify the integral on the left-hand side by setting \(u = y(t)\) and \(du = dy(t)\): \[\int \frac{1}{u} du = \int \frac{h(t)}{g(t)} dt\]
04

Solve the integrals

Solve the integrals on both sides of the equation: \[ln(|y(t)|) +C_1 = \int \frac{h(t)}{g(t)} dt + C_2\] To simplify and combine the constants, we can rewrite the equation as: \[ln(|y(t)|) = \int \frac{h(t)}{g(t)} dt + C\] where \(C = C_2 - C_1\).
05

Solve for \(y(t)\)

To solve for \(y(t)\), we can take the exponent of both sides of the equation: \[|y(t)| = e^{\int \frac{h(t)}{g(t)} dt + C}\] Since \(C\) is also an arbitrary constant, we can combine it with the constant of integration in the exponent: \[|y(t)| = e^{C} \cdot e^{\int \frac{h(t)}{g(t)} dt}\] Now, remove the absolute value by introducing another arbitrary constant, \(K\): \[y(t) = K \cdot e^{\int \frac{h(t)}{g(t)} dt}\] We have now obtained the general solution for a separable differential equation of the given form.

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Key Concepts

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

Integrating Factors
An integrating factor is a function used to multiply both sides of a differential equation to facilitate integration, often transforming a non-exact equation into an exact one. While the example at hand deals with separable differential equations directly, it's worth knowing how integrating factors might play a role in more complex scenarios. For instance, if one encounters a linear first-order differential equation of the form \(y' + p(t)y = g(t)\), an integrating factor, typically \(e^{\int p(t) dt}\), is used to enable the product rule of integration. Once multiplied, the left-hand side of the equation becomes the derivative of the product of the integrating factor and \(y(t)\), making it possible to integrate both sides with ease.
Variable Separation
Variable separation is a technique for solving first-order separable differential equations where variables related to \(y\) are isolated on one side and those related to \(t\) on the other. As seen in the solution steps, separate the functions involving \(y\) and \(t\), and then integrate both sides. For the given exercise, this step is crucial as it separates variables to prepare for individual integration. Remember, this method can only be applied when a differential equation is indeed separable, which is the case for equations of the form \(g(t) y'(t) = h(t)\).
Exponential Function Integration
Exponential function integration is particularly useful in solving differential equations involving growth and decay processes. In our exercise, after separating variables and integrating, we introduce the natural logarithm and exponentiate to solve for \(y(t)\). The relationship between logarithms and exponentials simplifies the process of solving for the variable \(y\). Integrating the exponential function typically yields an expression involving the natural logarithm, as shown in step 4 of the solution, and exponentiating both sides of the equation effectively 'undoes' the logarithm, bringing us closer to the solution for \(y(t)\).
Arbitrary Constants
In the process of integrating differential equations, we introduce arbitrary constants (\(C_1\), \(C_2\), etc.), which are essential in forming the general solution. These constants arise because indefinite integrals are not unique; adding a constant to any function's integral will yield another valid integral. After integration in the exercise, we combine constants from both sides to form a single arbitrary constant \(C\). Finally, by exponentiating and recognizing that \(e^C\) is also arbitrary (as it could be any positive number), we introduce a new arbitrary constant \(K\), encapsulating the exponential of any added constant and catering to the positive or negative values of \(y(t)\). Understanding the role of arbitrary constants is crucial as they ultimately determine the particular solution to a differential equation given initial conditions.

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Most popular questions from this chapter

The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for \(t \geq 0,\) graph the solution, and determine the first month in which the loan balance is zero. $$ B^{\prime}(t)=0.005 B-500, B(0)=50,000 $$

Motion in a gravitational field An object is fired vertically upward with an initial velocity \(v(0)=v_{0}\) from an initial position \(s(0)=s_{0}\) . a. For the following values of \(v_{0}\) and \(s_{0},\) find the position and velocity functions for all times at which the object is above the ground \((s=0).\) b. Find the time at which the highest point of the trajectory is reached and the height of the object at that time. $$v_{0}=29.4 \mathrm{m} / \mathrm{s}, s_{0}=30 \mathrm{m}$$

Equilibrium solutions \(A\) differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on y. The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=\sin y$$

Analytical solution of the predator-prey equations The solution of the predator-prey equations $$x^{\prime}(t)=-a x+b x y, y^{\prime}(t)=c y-d x y$$ can be viewed as parametric equations that describe the solution curves. Assume \(a, b, c,\) and \(d\) are positive constants and consider solutions in the first quadrant. a. Recalling that \(\frac{d y}{d x}=\frac{y^{\prime}(t)}{x^{\prime}(t)},\) divide the first equation by the second equation to obtain a separable differential equation in terms of \(x\) and \(y\) b. Show that the general solution can be written in the implicit form \(e^{d x+b y}=C x^{r} y^{a},\) where \(C\) is an arbitrary constant. c. Let \(a=0.8, b=0.4, c=0.9,\) and \(d=0.3 .\) Plot the solution curves for \(C=1.5,2,\) and \(2.5,\) and confirm that they are, in fact, closed curves. Use the graphing window \([0,9] \times[0,9]\)

Solving initial value problems Solve the following initial value problems. $$y^{\prime}(t)=t e^{t}, y(0)=-1$$
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