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Chapter 8: Problem 10

Find the general solution of the following equations. $$v^{\prime}(y)-\frac{v}{2}=14$$

Short Answer

Expert verified
Question: Find the general solution of the first-order linear ODE \(v^{\prime}(y)-\frac{v}{2}=14\). Answer: The general solution for the given ODE is \(v(y) = -28 + Ce^{\frac{1}{2}y}\).

Step by step solution

01

Identify the form of the ODE

The ODE is given in the form of a first-order linear ODE, which can be generally represented as: $$\frac{dv}{dy} + P(y)v = Q(y)$$ For our problem, we have: $$P(y) = -\frac{1}{2}$$ $$Q(y) = 14$$
02

Calculate the integrating factor

To find the integrating factor, we need to compute the exponential of the integral of P(y) with respect to y: $$I(y) = e^{\int P(y) dy}= e^{\int -\frac{1}{2} dy}$$ Integrating, we get: $$I(y) = e^{-\frac{1}{2}y}$$ Now multiply the ODE by the integrating factor: $$e^{-\frac{1}{2}y}(v^{\prime} - \frac{v}{2})= 14 e^{-\frac{1}{2}y}$$
03

Simplify and integrate both sides

Notice that the left-hand side is now the derivative of the product of v and the integrating factor, so we have: $$\frac{d}{dy}(ve^{-\frac{1}{2}y}) = 14 e^{-\frac{1}{2}y}$$ Now, integrate both sides with respect to y: $$\int \frac{d}{dy}(ve^{-\frac{1}{2}y}) dy = \int 14 e^{-\frac{1}{2}y} dy$$ This simplifies to: $$ve^{-\frac{1}{2}y} = -28e^{-\frac{1}{2}y} + C$$
04

Solve for the general solution

Now that we have integrated both sides, we need to solve for v. We can do this by multiplying both sides by the inverse of the integrating factor, which is \(e^{\frac{1}{2}y}\). This gives us: $$v(y) = -28 + Ce^{\frac{1}{2}y}$$ This is the general solution for the given ODE: $$v(y) = -28 + Ce^{\frac{1}{2}y}$$

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Most popular questions from this chapter

A differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(2-y)$$

Use the method outlined in Exercise 43 to solve the following Bernoulli equations. a. \(y^{\prime}(t)+y=2 y^{2}\) b. \(y^{\prime}(t)-2 y=3 y^{-1}\) c. \(y^{\prime}(t)+y=\sqrt{y}\)

Analysis of a separable equation Consider the differential equation \(y y^{\prime}(t)=\frac{1}{2} e^{t}+t\) and carry out the following analysis. a. Find the general solution of the equation and express it explicitly as a function of \(t\) in two cases: \(y>0\) and \(y < 0\) b. Find the solutions that satisfy the initial conditions \(y(-1)=1\) and \(y(-1)=2\) c. Graph the solutions in part (b) and describe their behavior as \(t\) increases. d. Find the solutions that satisfy the initial conditions \(y(-1)=-1\) and \(y(-1)=-2\) e. Graph the solutions in part (d) and describe their behavior as \(t\) increases.

Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1)\), the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\) c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\)

Write a logistic equation with the following parameter values. Then solve the initial value problem and graph the solution. Let \(r\) be the natural growth rate, \(K\) the carrying capacity, and \(P_{0}\) the initial population. $$r=0.2, K=300, P_{0}=50$$
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