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Chapter 9: Problem 34

Solving initial value problems Solve the following initial value problems. $$y^{\prime}(t)=\sin t+\cos 2 t, y(0)=4$$

Short Answer

Expert verified
Based on the solution above, determine the value of y(t) at t = 蟺/2. Using the expression for y(t), we substitute t = 蟺/2: $$y(\frac{\pi}{2}) = -\cos (\frac{\pi}{2}) + \frac{1}{2} \sin (2 \cdot \frac{\pi}{2}) + 5$$ From trigonometry, we know that $$\cos (\frac{\pi}{2}) = 0$$ and $$\sin(\pi) = 0$$. Thus, we have: $$y(\frac{\pi}{2}) = -0 + \frac{1}{2} \cdot 0 + 5$$ $$y(\frac{\pi}{2}) = 5$$

Step by step solution

01

Integrate the equation to find y(t)

First, let's find the function y(t) by integrating both sides of the equation with respect to t. $$\int y'(t) \, dt = \int (\sin t + \cos 2t) \, dt$$The integral of the left side is simply the function y(t). On the right side, we have the integrals of the sine and cosine functions: $$\int \sin t \, dt + \int \cos 2t \, dt$$. Now we need to integrate these functions. To integrate $$\sin t$$, we have $$\int \sin t \, dt = -\cos t$$ Next, integrating $$\cos 2t$$ involves a substitution. Let $$u = 2t$$, so $$du/dt = 2$$. Now, we have $$\int \cos 2t \, dt = \frac{1}{2} \int \cos u \, du$$ Integrating $$\cos u$$, we have $$\frac{1}{2} \int \cos u \, du = \frac{1}{2}\sin u$$ Now, substitute back for u, we get $$\frac{1}{2} \sin (2t)$$ Adding the results of the integrations, we get $$y(t) = -\cos t + \frac{1}{2}\sin (2t) + C$$ where C is the constant of integration.
02

Use the initial condition to find the unknown constant of integration

Now we need to find the value of C using the initial condition $$y(0) = 4$$. Substitute t = 0 into the equation for y(t), we get $$y(0) = -\cos (0) + \frac{1}{2}\sin 0 + C$$ From trigonometry, we know that $$\cos(0) = 1$$ and $$\sin(0) = 0$$. So, we have $$4 = -1 + 0 + C$$ Solving for C, we get $$C = 5$$
03

Combine the results to obtain the solution of the initial value problem

Now we have all the pieces to find the solution to the initial value problem. The final solution for y(t) is $$y(t) = -\cos t + \frac{1}{2}\sin (2t) + 5$$

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Key Concepts

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

Integrating Trigonometric Functions
Integrating trigonometric functions is a fundamental skill when solving differential equations involving terms like \(\sin t\) and \(\cos t\).

While integrating, the key is to remember the basic integrals: \(\int \sin t \, dt = -\cos t\) and \(\int \cos t \, dt = \sin t\). However, when the trigonometric function has a coefficient within the argument, such as \(\cos 2t\), we need to use a substitution method for integration.

For example, let's consider the term \(\cos 2t\). We set \(u = 2t\) which leads to \(du = 2 dt\), and thus \(dt = \frac{1}{2}du\). Substituting into the integral, we get \(\frac{1}{2} \int \cos u \, du\), which simplifies to \(\frac{1}{2}\sin u\). When we replace \(u\) with \(2t\), we obtain \(\frac{1}{2} \sin(2t)\).

This method allows us to break down more complex integral expressions into simpler ones that can be easily integrated using basic knowledge of trigonometry.
Constant of Integration
When we integrate any function, we introduce an arbitrary constant, known as the constant of integration, denoted by \(C\). This constant represents the infinite number of antiderivatives a function can have, all of which differ by a constant value.

For instance, after integrating \(\int (\sin t + \cos 2t) \, dt\), we obtain \(y(t) = -\cos t + \frac{1}{2}\sin (2t) + C\). The \(C\) captures all possible vertical shifts of the antiderivative's graph. To determine its value, we use additional information, often called an initial condition.

In the given exercise, we use the fact that \(y(0) = 4\) to find the specific value of \(C\). When \(t = 0\), the equation simplifies using basic trigonometric identities, allowing us to solve for \(C\) uniquely. By substituting the known initial values into the integrated function, we can calculate the exact value of this constant, making our general solution a particular one that satisfies all given conditions.
Solving Differential Equations
Solving differential equations often involves finding a function that satisfies a given relation between its derivatives. In the case of simple first-order differential equations, we integrate both sides with respect to the independent variable, which is typically time or space.

The process involves recognizing the type of differential equation and then applying the appropriate method 鈥� in this case, direct integration. After the integration, we obtain a general solution that includes a constant of integration.

However, in order to determine a unique solution for a differential equation, an initial value or boundary condition is required. This is known as an initial value problem. By applying the given condition, as we did in Step 2 of our example, we can determine the precise value for the constant of integration, leading to the specific solution that satisfies both the differential equation and the initial condition: \(y(t) = -\cos t + \frac{1}{2}\sin (2t) + 5\).

This process ties together the need to not only integrate trigonometric functions but also to precisely understand the significance of the constant of integration to wholly solve the differential equation at hand.

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Most popular questions from this chapter

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 \([0,9] \times[0,9]\)

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. $$\left(t^{2}+1\right) y^{\prime}(t)+2 t y=3 t^{2}, y(2)=8$$

In each case, carry out the following steps. a. Identify which equation corresponds to the predator and which corresponds to the prey. b. Find the lines along which \(x^{\prime}(t)=0 .\) Find the lines along which \(y^{\prime}(t)=0\) c. Find the equilibrium points for the system. d. Identify the four regions in the first quadrant of the xy-plane in which \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. e. Sketch a representative solution curve in the xy-plane and indicate the direction in which the solution evolves. $$x^{\prime}(t)=2 x-4 x y, y^{\prime}(t)=-y+2 x y$$

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.

The amount of drug in the blood of a patient (in milligrams) administered via an intravenous line is governed by the initial value problem \(y^{\prime}(t)=-0.02 y+3\) \(y(0)=0,\) where \(t\) is measured in hours. a. Find and graph the solution of the initial value problem. b. What is the steady-state level of the drug? c. When does the drug level reach \(90 \%\) of the steady-state value?
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