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Chapter 9: Problem 28

Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=\frac{y+3}{5 t+6}, y(2)=0$$

Short Answer

Expert verified
The solution to the initial value problem is \(y(t) = \frac{3}{(16)^{\frac{1}{5}}}(5t+6)^{\frac{1}{5}} - 3\).

Step by step solution

01

Check if the equation is separable

To check if the given equation is separable, we rewrite it as: $$y'(t) = \frac{y+3}{5t+6}$$ Now, let's see if we can separate the two variables, y and t: $$\frac{dy}{dt} = \frac{y+3}{5t+6}$$ $$dy = \frac{y+3}{5t+6}dt$$ Yes, we were able to separate the variables y and t. Thus, the equation is separable.
02

Integrate both sides

Now that the equation is separable, we can integrate both sides with respect to their respective variables: $$\int \frac{dy}{y+3} = \int \frac{dt}{5t+6}$$
03

Use integration techniques to solve the integrals

For the left side, we already have an elementary function to integrate: \(\ln|y+3|\). Now, for the right side, we can make a substitution to find the integral: Let \(u = 5t+6\), so \(\frac{du}{dt} = 5\). Therefore, \(dt=\frac{du}{5}\). Now, we substitute u in the right integral: $$\int \frac{dy}{y+3} = \int \frac{1}{u}\frac{du}{5} = \frac{1}{5}\int \frac{du}{u}$$ So, the equation after integrating both sides is: $$\ln|y+3| = \frac{1}{5}\ln|u| + C_1$$
04

Solve for y and apply the initial condition

First, let's resubstitute the value of u into the equation: $$\ln|y+3| = \frac{1}{5}\ln|5t+6| + C_1$$ Now, exponentiate both sides to get rid of the natural logarithm: $$y+3 = e^{\frac{1}{5}\ln|5t+6|+C_1}$$ Remove the exponential and logarithm, we have $$y+3 = e^{C_1} (5t+6)^{\frac{1}{5}}$$ Now, let's denote \(e^{C_1}\) as a new constant: \(C_2\): $$y(t) = C_2(5t+6)^{\frac{1}{5}} -3$$ Now, apply the initial condition \(y(2) = 0\): $$0 = C_2(5(2) +6)^{\frac{1}{5}} - 3$$ Solve for \(C_2\) to obtain: $$C_2 = \frac{3}{(16)^{\frac{1}{5}}}$$
05

Write the final solution

Now, substitute the value of \(C_2\) into the equation we derived for y(t): $$y(t) = \frac{3}{(16)^{\frac{1}{5}}}(5t+6)^{\frac{1}{5}} - 3$$ This is the solution to the initial value problem.

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a class of equations that can be rearranged to allow the integration of two variables independently, essentially 'separating' the variables. The key strategy is to isolate all instances of one variable and its differential on one side of the equation, and the other variable and its differential on the other side.

For example, a standard separable equation appears as \( \frac{dy}{dx} = g(x)f(y) \), which can be reorganized to \( \frac{dy}{f(y)} = g(x)dx \). Once separated, you can integrate both sides independently: \( \int \frac{1}{f(y)}dy = \int g(x)dx \). Applying this technique to our initial value problem, \( y'(t) = \frac{y+3}{5t+6} \) and with \( y(2) = 0 \) as the initial condition, the equation can easily be turned into a format that can be integrated with respect to \( y \) and \( t \) on each side.
Integrating Factors
An integrating factor is a function that is often used to solve linear non-separable differential equations. However, for the sake of this explanation, focus will be on how this concept is related to simplifying equations for integration. In some cases, rearranging an equation doesn't immediately reveal a form that can be integrated; an integrating factor can be multiplied with the equation to make it integrable.

The general idea is to find a function, often denoted as \( \mu(x) \) or \( \mu(t) \) in our initial value problem, such that when multiplied by our equation, the left side becomes the derivative of a product of two functions. In the context of our problem, the equation was already in a separable form, so we didn't need to use an integrating factor. However, it is important to recognize that integrating factors can simplify certain equations by turning them into a form where the standard integration techniques can be employed.
Exponential Functions
Exponential functions play a vital role in solving differential equations, especially when dealing with logarithmic separations that arise after integration. The natural logarithm and exponential function are inverses of each other, and this property is particularly useful for eliminating logarithmic terms after integrating both sides of an equation.

In our initial value problem, after finding the natural logarithm of both sides, we utilized the property that \( e^{\ln x} = x \) to rewrite the solution in terms of the original variables. This resulted in the exponential term \( e^{\frac{1}{5}\ln|5t+6|+C_1} \), which could be simplified using exponent rules to \( (5t+6)^{\frac{1}{5}} \) multiplied by the new constant \( C_2 = e^{C_1} \). Exponential functions are also cornerstone functions in modeling growth and decay processes in various disciplines.

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Most popular questions from this chapter

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t .\) Use this idea to solve the following initial value problems. $$t^{3} y^{\prime}(t)+3 t^{2} y=\frac{1+t}{t}, y(1)=6$$

Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. Chemical rate equations Consider the chemical rate equations \(y^{\prime}(t)=-k y(t)\) and \(y^{\prime}(t)=-k y^{2}(t),\) where \(y(t)\) is the concentration of the compound for \(t \geq 0,\) and \(k>0\) is a constant that determines the speed of the reaction. Assume the initial concentration of the compound is \(y(0)=y_{0}>0\). a. Let \(k=0.3\) and make a sketch of the direction fields for both equations. What is the equilibrium solution in both cases? b. According to the direction fields, which reaction approaches its equilibrium solution faster?

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. $$B^{\prime}(t)=0.0075 B-1500, B(0)=100,000$$

Direction field analysis Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a\), then the solution increases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a\), then the solution decreases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

General Gompertz solution Solve the initial value problem $$M^{\prime}(t)=-r M \ln \left(\frac{M}{K}\right), M(0)=M_{0}$$ with arbitrary positive values of \(r, K,\) and \(M_{0^{\prime}}\)
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