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Chapter 8: Problem 4

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

Short Answer

Expert verified
In this exercise, we were given a separable differential equation of the form \(g(y)y'(t)=h(t)\). To solve this equation, we first verified that it is separable. Since both \(g(y)\) and \(h(t)\) are functions of only one variable, the equation is separable. Next, we rearranged the equation to isolate terms with \(y\) on one side and terms with \(t\) on the other side. Then, we integrated both sides of the equation with respect to their respective variables. After integrating, we obtained a general solution in the form of an implicit equation involving \(y\) and \(t\). Depending on whether we could solve for \(y(t)\) explicitly or not, the general solution was either given in an explicit form as \(y(t) = f(t) + C\) or left in its implicit form. This general solution represents the family of all possible solutions to the given separable differential equation, with the constant of integration \(C\) determined by any given initial conditions.

Step by step solution

01

Verify the equation is separable

In this case, we are given a differential equation \(g(y)y'(t)=h(t)\). Both \(g(y)\) and \(h(t)\) are functions of only one variable, so the equation is already separable.
02

Rearrange the equation

To separate the variables, we will divide both sides by \(g(y)\) so that we have all terms involving \(y\) on the left side and all terms involving \(t\) on the right side: $$\frac{y'(t)}{g(y)} = \frac{h(t)}{g(y)}.$$ We can rewrite this differential equation as $$\frac{dy}{dt} \cdot \frac{1}{g(y)} = \frac{h(t)}{g(y)}.$$
03

Integrate both sides

Now, we will integrate both sides of the equation with respect to their respective variables. For the left side, we integrate with respect to \(y\), and for the right side, we integrate with respect to \(t\): $$\int \frac{1}{g(y)} dy = \int \frac{h(t)}{g(y)} dt.$$
04

Solve for y(t)

After integrating both sides, we will get a general solution in the form of an implicit equation involving \(y\) and \(t\). We may or may not be able to solve this equation explicitly for \(y(t)\). If we can find an explicit solution, we write \(y(t) = f(t) + C\), where \(f(t)\) is the function of \(t\) found by integrating, and \(C\) is the constant of integration. If we cannot find an explicit solution, we will leave the equation in its implicit form.
05

Provide the general solution

The general solution will either be the explicit solution of the form \(y(t) = f(t) + C\) or the implicit solution, depending on whether we were able to solve for \(y(t)\) in Step 4. This solution represents the family of all possible solutions to the given separable differential equation, with the constant of integration \(C\) determined by any given initial conditions.

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Key Concepts

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

Integration Techniques
Integration is a crucial technique for solving differential equations. In the context of separable differential equations, we use integration to find the antiderivatives of functions associated with each variable. To do this, we separate the variables first, ensuring that one side of the equation only involves the function of one variable, and the opposite side involves the other. This separation allows us to integrate both sides individually:
  • On the left side, integrate the expression involving the function of the dependent variable, typically denoted as \(y\).
  • On the right side, integrate with respect to the independent variable, usually \(t\).
The process of integration helps to transform a problem expressed with derivatives into one expressed with functions, making it easier to solve. Clearly understanding how to apply integration methods will simplify solving differential equations significantly, even if one side ends up more complicated than the other. When dealing with complex functions, proficiency in common integration techniques, such as substitution or integration by parts, is highly beneficial.
Differential Equations
A differential equation relates a function with its derivatives, showing how the function changes over time or space. In this case, we are focusing on separable differential equations. These are specifically structured so that all terms involving the dependent variable can be separated from those involving the independent variable. The general form of a separable equation given in this context is \(g(y) y'(t) = h(t)\). This behavior allows us to handle each variable separately, simplifying the overall equation.
Differential equations can describe many real-world phenomena, like population growth, heat conduction, and motion dynamics, making them invaluable in various fields such as physics, biology, and engineering. By breaking down the equation into manageable parts, we can identify not just a singular solution, but often a family of solutions with different constants of integration. This procedure might seem daunting at first, but understanding how these equations inherently behave provides a powerful tool for analyzing dynamic systems.
Implicit Solutions
Implicit solutions arise when it's not possible to solve explicitly for a dependent variable from an implicit equation after integration. These solutions are just as valid as explicit ones, providing a relationship between the dependent and independent variables.
  • An implicit solution doesn't isolate the dependent variable (\(y\)) in terms of the independent variable (\(t\)), but rather presents a general equation involving both variables.
  • Example: After integration, the solution might be \( F(y, t) = 0 \), where \(F\) is some function defining the relationship between \(y\) and \(t\).
Such solutions fit scenarios where separating out \(y(t)\) isn't feasible. Importantly, implicit solutions may also cater to complex or multi-valued situations, where an explicit solution could be overly restrictive or complicated to derive. Ultimately, it's critical to accept implicit solutions as perfectly functional representations within the larger scope of solution sets for differential equations.

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Most popular questions from this chapter

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

Consider a loan repayment plan described by the initial value problem $$B^{\prime}(t)=0.03 B-600, \quad B(0)=40,000$$ where the amount borrowed is \(B(0)=\$ 40,000,\) the monthly payments are \(\$ 600,\) and \(B(t)\) is the unpaid balance in the loan. a. Find the solution of the initial value problem and explain why \(B\) is an increasing function. b. What is the most that you can borrow under the terms of this loan without going further into debt each month? c. Now consider the more general loan repayment plan described by the initial value problem $$B^{\prime}(t)=r B-m, \quad B(0)=B_{0}$$ where \(r>0\) reflects the interest rate, \(m>0\) is the monthly payment, and \(B_{0}>0\) is the amount borrowed. In terms of \(m\) and \(r,\) what is the maximum amount \(B_{0}\) that can be borrowed without going further into debt each month?

Determine whether the following equations are separable. If so, solve the initial value problem. $$\sec x y^{\prime}(x)=y^{3}, y(0)=3$$

Explain how the growth rate function determines the solution of a population model.

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 that \(a, b, c,\) 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 \(e^{d x+b y}=C x^{c} y^{a},\) 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 \([0,9] \times[0,9]\)
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