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Chapter 9: Problem 32

Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$(\sec t) y^{\prime}(t)=1, y(0)=1$$

Short Answer

Expert verified
Answer: The solution to the initial value problem is \(y(t) = \sin t + 1\).

Step by step solution

01

Determine if the equation is separable

A separable equation is an equation that can be written in the form of \(g(y) y^{\prime}(t) = h(t)\). To find out if the given differential equation is separable, let's rewrite the given equation as: $$y^{\prime}(t) = \frac{1}{\sec t}$$ Since \(\sec t = \frac{1}{\cos t}\), we can also write it as: $$y^{\prime}(t) = \cos t$$ Since the given differential equation can be written in the form of \(y^{\prime}(t) = h(t)\), it is a separable equation.
02

Separate the variables and integrate

The differential equation can be written as: $$\frac{dy}{dt} = \cos t$$ Now, separate the variables: $$dy = \cos t \ dt$$ Now integrate both sides: $$\int dy = \int \cos t \ dt$$ $$\Rightarrow y(t) = \int \cos t \ dt$$
03

Find the general solution of the differential equation

Now, to find the integral of \(\cos t\), use the antiderivative: $$y(t) = \int \cos t \ dt = \sin t + C$$ where \(C\) is the constant of integration.
04

Apply the initial condition \(y(0)=1\)

The initial condition states that \(y(0)=1\). Use this condition to find the constant \(C\): $$y(0) = \sin(0) + C = 1$$ $$\Rightarrow C = 1$$
05

Write down the solution to the initial value problem

Now that we have found the constant \(C\), substitute it back into the general solution: $$y(t) = \sin t + 1$$ So, the solution to the initial value problem is: $$y(t) = \sin t + 1$$

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Key Concepts

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

Initial Value Problem
An initial value problem (IVP) in mathematics is essentially about finding a solution to a differential equation, where some initial conditions are given. In simple terms, an IVP seeks to find a unique function that fits into a differential equation and also meets a specific condition at a given point.
Here is what you need to know about initial value problems:
  • The equation might look complicated, but the idea is straightforward: identify an equation that satisfies both the differential equation and the initial condition.
  • The initial condition is usually given in the form of a specific point, such as \( y(a) = b \).For our problem, the initial condition is \( y(0) = 1 \), which means when \( t=0 \), the function value \( y \) is 1.
  • Solving an IVP involves two steps: finding a general solution to the differential equation and then applying the initial condition to find the particular solution.
Integration
Integration is one of the core processes we use in calculus. It essentially means finding the integral or antiderivative of a function. When solving a differential equation, integration is used to obtain the original function from its derivative.
Here's what to keep in mind about integration:
  • The integration process helps in finding the general form of a function by reversing the process of differentiation. It is like unraveling a mystery.
  • In solving separable differential equations, once the variables are separated, each side is integrated independently. For example, in the problem, the equation was separated to \( dy = \cos t \, dt \), and then both sides were integrated to \( \int dy = \int \cos t \, dt \).
  • The result of integration includes the constant of integration (usually denoted by \( C \)), which represents the infinite set of possible solutions.
Antiderivative
An antiderivative is a function that reverses differentiation. It is the opposite of taking a derivative. When solving differential equations, finding an antiderivative is key to getting back to the original function from its rate of change.
Here's how understanding antiderivatives aids in solving separable differential equations:
  • Antiderivatives determine the family of functions that have a given function as their derivative. For instance, the antiderivative of \( \cos t \) is \( \sin t \), because the derivative of \( \sin t \) returns \( \cos t \).
  • Finding the antiderivative is crucial in solving separable equations because once you find this function, you are a step closer to identifying the specific solution.
  • Each antiderivative solution includes a constant of integration (\( C \)) because derivatives of constants are zero, and thus constants do not appear in the derived equation.

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Most popular questions from this chapter

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 \(2 x^{2}+y^{2}=a^{2}\).

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

Properties of stirred tank solutions a. Show that for general positive values of \(R, V, C_{i},\) and \(m_{0},\) the solution of the initial value problem $$m^{\prime}(t)=-\frac{R}{V} m(t)+C_{i} R, \quad m(0)=m_{0}$$is \(m(t)=\left(m_{0}-C_{i} V\right) e^{-R t / V}+C_{i} V\) b. Verify that \(m(0)=m_{0}\) c. Evaluate \(\lim _{t \rightarrow \infty} m(t)\) and give a physical interpretation of the result. d. Suppose \(m_{0}\) and \(V\) are fixed. Describe the effect of increasing \(R\) on the graph of the solution.

Convergence of Euler's method Suppose Euler's method is applied to the initial value problem \(y^{\prime}(t)=a y, y(0)=1,\) which has the exact solution \(y(t)=e^{a t} .\) For this exercise, let \(h\) denote the time step (rather than \(\Delta t\) ). The grid points are then given by \(t_{k}=k h .\) We let \(u_{k}\) be the Euler approximation to the exact solution \(y\left(t_{k}\right),\) for \(k=0,1,2, \ldots\). a. Show that Euler's method applied to this problem can be written \(u_{0}=1, u_{k+1}=(1+a h) u_{k},\) for \(k=0,1,2, \ldots\). b. Show by substitution that \(u_{k}=(1+a h)^{k}\) is a solution of the equations in part (a), for \(k=0,1,2, \ldots\). c. Recall from Section 4.7 that \(\lim _{h \rightarrow 0}(1+a h)^{1 / h}=e^{a} .\) Use this fact to show that as the time step goes to zero \((h \rightarrow 0,\) with \(\left.t_{k}=k h \text { fixed }\right),\) the approximations given by Euler's method approach the exact solution of the initial value problem; that is, \(\lim _{h \rightarrow 0} u_{k}=\lim _{h \rightarrow 0}(1+a h)^{k}=y\left(t_{k}\right)=e^{a t_{k}}\).

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 \(t \geq 0\).
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