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Chapter 9: Problem 16

Solve the initial value problems in Exercises \(15-20\) $$t \frac{d y}{d t}+2 y=t^{3}, \quad t>0, \quad y(2)=1$$

Short Answer

Expert verified
The solution is \( y(t) = \frac{t^3}{5} - \frac{12}{5t^2} \).

Step by step solution

01

Rewrite the Differential Equation

Given the differential equation: \( t \frac{d y}{d t} + 2y = t^3 \), let's rewrite it in the standard linear form: \( \frac{d y}{d t} + \frac{2}{t} y = t^2 \). This is a first-order linear differential equation.
02

Find the Integrating Factor

Identify the integrating factor \( \mu(t) \) for the differential equation. It is found using the formula \( \mu(t) = e^{\int P(t) \, dt} \), where \( P(t) = \frac{2}{t} \). Thus, \( \mu(t) = e^{\int \frac{2}{t} \, dt} = e^{2 \ln |t|} = t^2 \).
03

Multiply through by the Integrating Factor

Multiply each term of the equation \( \frac{d y}{d t} + \frac{2}{t} y = t^2 \) by the integrating factor \( t^2 \). This gives us: \( t^2 \frac{d y}{d t} + 2t y = t^4 \).
04

Solve the Resulting Equation

Recognize that the left side \( t^2 \frac{d y}{d t} + 2t y = \frac{d}{dt}(t^2 y) \). So the equation simplifies to: \( \frac{d}{dt}(t^2 y) = t^4 \).
05

Integrate Both Sides

Integrate both sides with respect to \( t \): \( \int \frac{d}{dt}(t^2 y) \, dt = \int t^4 \, dt \). This gives \( t^2 y = \frac{t^5}{5} + C \), where \( C \) is the integration constant.
06

Solve for y(t)

Divide both sides of \( t^2 y = \frac{t^5}{5} + C \) by \( t^2 \) to solve for \( y(t) \): \( y(t) = \frac{t^3}{5} + \frac{C}{t^2} \).
07

Apply Initial Condition

Use the initial condition \( y(2) = 1 \) to solve for \( C \). Substitute into the equation: \( 1 = \frac{2^3}{5} + \frac{C}{2^2} \). Calculate \( \frac{8}{5} + \frac{C}{4} = 1 \). Simplify to find \( \frac{C}{4} = 1 - \frac{8}{5} = \frac{-3}{5} \). Hence, \( C = -\frac{12}{5} \).
08

Write the Final Solution

Substitute the value of \( C \) back into the expression for \( y(t) \): \( y(t) = \frac{t^3}{5} - \frac{12}{5t^2} \). This is the particular solution satisfying the initial condition.

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Key Concepts

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

Integrating factor
The integrating factor is a crucial tool when solving first-order linear differential equations. The idea is to multiply the entire equation by this special function, which can unify the disparate elements of the equation into an easily integrable form. Here's a simple breakdown of the concept:
  • The differential equation is first rewritten in the standard linear form: \( \frac{d y}{d t} + P(t) y = Q(t) \).
  • The integrating factor \( \mu(t) \) is determined using the formula \( \mu(t) = e^{\int P(t) \, dt} \).
  • This magical function transforms the left side of the equation into the derivative of a product, \( \frac{d}{dt}(\mu(t)y) \). This transformation simplifies the process of solving the equation.
In our problem, the function \( P(t) \) was \( \frac{2}{t} \). Calculating the integrating factor gives us \( \mu(t) = t^2 \). Using this \( \mu(t) \), each term of the differential equation: \( \frac{d y}{d t} + \frac{2}{t} y = t^2 \) is multiplied by \( t^2 \), achieving the simplified integrable form.
Initial value problem
An initial value problem (IVP) is a type of differential equation paired with specific conditions at a certain point. This specific condition is called the "initial condition," which helps find a unique solution to the differential equation. Here's why it is important:
  • An initial value problem is presented as a differential equation with a condition such as \( y(t_0) = y_0 \).
  • The initial condition is critical because it allows us to solve for any constant that arises during the solution process.
  • Solving an IVP ensures the solution conforms to a specific real-world scenario described by the initial condition.
In our exercise, the initial condition given was \( y(2) = 1 \). After integrating and deforming our equation to \( y(t) = \frac{t^3}{5} + \frac{C}{t^2} \), this condition was applied to determine the constant \( C \). Solving \( 1 = \frac{2^3}{5} + \frac{C}{4} \), which led to \( C = -\frac{12}{5} \). This made the solution specific to the initial scenario.
Solution of differential equations
Solving a differential equation means finding a function, or a set of functions, that satisfy the equation. Especially with first-order linear differential equations, several steps typically unfold in the solution process:
  • First, the differential equation is rearranged into standard form.
  • Next, we determine an integrating factor to facilitate integration across both sides of the equation.
  • Once both sides are integrated, a solution formula involving an integration constant \( C \) typically emerges.
  • The inclusion of an initial condition allows for determining \( C \), yielding a specific solution fitting a real situation.
Our example led to such a solution: once \( t^2 \) and \( y(t) \) were effectively combined and integrated, the function \( y(t) = \frac{t^3}{5} - \frac{12}{5t^2} \) was obtained. This function solved both the differential equation and fulfilled the initial condition \( y(2) = 1 \). It shows a unique, particular solution fitting the specific context provided by the problem.

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Most popular questions from this chapter

Obtain a slope field and add to it graphs of the solution curves passing through the given points. \(y^{\prime}=2(y-4)\) with a. \((0,1) \quad\) b. \((0,4) \quad\) c. \((0,5)\)

Solve the initial value problems in Exercises \(15-20\) $$\theta \frac{d y}{d \theta}+y=\sin \theta, \quad \theta>0, \quad y(\pi / 2)=1$$

Use the Euler method with \(d x=1 / 3\) to estimate \(y(2)\) if \(y^{\prime}=\) \(x \sin y\) and \(y(0)=1 .\) What is the exact value of \(y(2) ?\)

An economic model Consider the following economic model. Let \(P\) be the price of a single item on the market. Let \(Q\) be the quantity of the item available on the market. Both \(P\) and \(Q\) are functions of time. If one considers price and quantity as two inter- acting species, the following model might be proposed: $$ \begin{aligned} \frac{d P}{d t} &=a P\left(\frac{b}{Q}-P\right) \\ \frac{d Q}{d t} &=c Q(f P-Q) \end{aligned} $$ where \(a, b, c,\) and \(f\) are positive constants. Justify and discuss the adequacy of the model. $$ \begin{array}{l}{\text { a. If } a=1, b=20,000, c=1, \text { and } f=30 \text { , find the equilibrium }} \\ {\text { points of this system. If possible, classify each equilibrium }} \\ {\text { point with respect to its stability. If a point cannot be }} \\ {\text { readily classified, give some explanation. }}\\\\{\text { b. Perform a graphical stability analysis to determine what will }} \\ {\text { happen to the levels of } P \text { and } Q \text { as time increases. }} \\ {\text { c. Give an economic interpretation of the curves that determine }} \\ {\text { the equilibrium points. }}\end{array} $$

Use a CAS to explore graphically each of the differential equations. Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y\)-window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b].\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\)-interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error \((y\) (exact) \(-y\) (Euler)) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error. \(y^{\prime}=(\sin x)(\sin y), \quad y(0)=2 ; \quad-6 \leq x \leq 6, \quad-6 \leq y \leq 6;\) \(b=3 \pi / 2\)
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