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Chapter 9: Problem 46

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t)$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{3}{t} y(t)=1-2 t, y(2)=0$$

Short Answer

Expert verified
In this exercise, the initial value problem given to us is: $$y^{\prime}(t)+\frac{3}{t} y(t)=1-2 t$$ with the initial condition $$y(2)=0$$. After following the method of integrating factors and incorporating the initial condition, we found the solution to the problem: $$y(t) = \frac{1}{4}t - \frac{2}{5}t^2 + \frac{16}{5t^3}$$

Step by step solution

01

Calculate the integrating factor

To find the integrating factor p(t), we need to compute the integral of the coefficient of y(t), which is \(\frac{3}{t}\). Calculate the integral: $$p(t) = \exp\left(\int \frac{3}{t} dt\right)$$ Integrating, we get: $$p(t) = \exp(3 \ln(t)) = t^3$$
02

Multiply the given equation by the integrating factor

Now, multiply the given equation by the integrating factor p(t) = \(t^3\): $$t^3\left(y^{\prime}(t)+\frac{3}{t} y(t)\right)=t^3(1-2 t)$$
03

Rewrite the left-hand side as an exact derivative

Taking the derivative of the product (p(t)y(t)), we get the left side of the equation: $$\frac{d}{dt}(t^3y(t)) = t^3(1-2t)$$
04

Integrate both sides of the equation

Integrate both sides of the equation with respect to t: $$\int \frac{d}{dt}(t^3y(t)) dt=\int t^3(1-2 t) dt$$ This yields: $$t^3y(t) = \int t^3 - 2t^4 dt = \frac{1}{4}t^4 - \frac{2}{5}t^5 + C$$
05

Find the constant of integration using the initial condition

To find the constant of integration C, we use the initial condition y(2) = 0: $$0 = \frac{1}{4}(2)^4 - \frac{2}{5}(2)^5 + C$$ Solving for C, we get: $$C = \frac{16}{5}$$
06

Solve for y(t)

Plug the value of C back into the equation and solve for y(t): $$t^3y(t) = \frac{1}{4}t^4 - \frac{2}{5}t^5 + \frac{16}{5}$$ Divide both sides by \(t^3\) to obtain the solution for y(t): $$y(t) = \frac{1}{4}t - \frac{2}{5}t^2 + \frac{16}{5t^3}$$ Thus, the solution for the given initial value problem is: $$y(t) = \frac{1}{4}t - \frac{2}{5}t^2 + \frac{16}{5t^3}$$

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Key Concepts

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

Integrating factor
When dealing with first-order linear differential equations, the integrating factor method is a powerful technique. The integrating factor is a special function, often introduced as \( p(t) \), that helps simplify the differential equation into an easier form to solve.
To find this integrating factor, you calculate an exponential of the integral of the coefficient of \( y(t) \). In simple terms, if you have an equation like \( y'(t) + a(t)y(t) = f(t) \), the integrating factor \( p(t) \) is given by:
  • \( p(t) = \exp\left(\int a(t) \, dt \right) \)
The beauty of the integrating factor is that it transforms the left side of the differential equation into an exact derivative. This means that instead of solving a differential equation, we need to solve a new equation that involves integration, making it much simpler!
Initial value problem
An initial value problem is a type of problem in which the solution to a differential equation is required to satisfy a specific initial condition. These problems provide a way to find a unique solution to differential equations.
For example, in the exercise given, the equation \( y'(t) + \frac{3}{t}y(t) = 1 - 2t \) is equipped with an initial condition \( y(2) = 0 \). This initial condition specifies the value of the unknown function \( y(t) \) at a particular point, here \( t = 2 \).
By incorporating this initial condition into the general solution of the differential equation, we can solve for any constant of integration that arises, ensuring the solution fits the criteria given at \( t = 2 \). This approach ensures the solution is both accurate and specific to the problem statement.
Exact derivative
The concept of an exact derivative plays a crucial role in simplifying first-order linear differential equations. An expression is an exact derivative if it can be written as the derivative of a product of functions.
In the context of solving these equations, once the original differential equation is multiplied by the integrating factor, the left-hand side becomes an exact derivative. This transformation is what allows for a straightforward integration process.
For instance, after applying our integrating factor \( t^3 \) to the equation \( y'(t) + \frac{3}{t}y(t) = 1 - 2t \), the left side becomes \( \frac{d}{dt}(t^3y(t)) \). This is now an exact derivative, which means integrating becomes a matter of reversing this derivative to find \( t^3y(t) \).
By identifying and using the exact derivative form, we reduce a differential equation to a basic integration problem, a much simpler mathematical task.

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Most popular questions from this chapter

Solve the initial value problem and graph the solution. Let \(r\) be the natural growth rate, \(K\) the carrying capacity, and \(P_{\mathrm{o}}\) the initial population. $$r=0.4, K=5500, P_{0}=100$$

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. Chemical rate equations The reaction of certain chemical compounds can be modeled using a differential equation of the form \(y^{\prime}(t)=-k y^{n}(t),\) where \(y(t)\) is the concentration of the compound, for \(t \geq 0, k>0\) is a constant that determines the speed of the reaction, and \(n\) is a positive integer called the order of the reaction. Assume the initial concentration of the compound is \(y(0)=y_{0}>0\) a. Consider a first-order reaction \((n=1)\) and show that the solution of the initial value problem is \(y(t)=y_{0} e^{-k t}\) b. Consider a second-order reaction \((n=2)\) and show that the solution of the initial value problem is \(y(t)=\frac{y_{0}}{y_{0} k t+1}\) c. Let \(y_{0}=1\) and \(k=0.1 .\) Graph the first-order and secondorder solutions found in parts (a) and (b). Compare the two reactions.

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?

Solve the equation \(y^{\prime}(t)=k y+b\) in the case that \(k y+b<0\) and verify that the general solution is \(y(t)=C e^{k t}-\frac{b}{k}\)

Orthogonal trajectories 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 \(y^{2}=k x\) where \(k\) is an arbitrary constant. 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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