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Chapter 8: Problem 14

Solve the following initial value problems. $$u^{\prime}(x)=2 u+6, u(1)=6$$

Short Answer

Expert verified
Question: Determine the explicit solution for the function \(u(x)\) that satisfies the given initial value problem: \(u^{\prime}(x)=2 u(x) + 6,\) with the initial condition \(u(1)=6\). Answer: The explicit solution for \(u(x)\) is \(u(x) = 3 + \frac{3(e^2-1)}{e^{2x}}.\)

Step by step solution

01

Identify the differential equation and initial condition

The given differential equation is: $$u^{\prime}(x)=2 u(x) + 6,$$ and the initial condition is: $$u(1)=6.$$
02

Use the integration factor method to solve the ODE

First, we determine the integrating factor: $$\mu(x) = e^{\int 2 \mathrm{d}x} = e^{2x}.$$ Multiply both sides of the ODE by the integrating factor: $$e^{2x}u^{\prime}(x) = 2e^{2x}u(x) + 6e^{2x}.$$ Now, we notice that the left side of the above equation is the derivative of the product \(e^{2x}u(x)\). So, rewrite the equation as: $$(e^{2x}u(x))^{\prime} = 6e^{2x}.$$
03

Integrate both sides of the equation

Integrate both sides of the equation with respect to x: $$\int(e^{2x}u(x))^{\prime} \mathrm{d}x = \int 6e^{2x} \mathrm{d}x.$$ This results in: $$e^{2x}u(x) = 3e^{2x} + C,$$ where C is the constant of integration.
04

Solve for u(x)

Divide both sides of the equation by \(e^{2x}\) to find \(u(x)\): $$u(x) = 3 + \frac{C}{e^{2x}}.$$
05

Use the initial condition to find the constant of integration

Apply the initial condition \(u(1) = 6\) to determine C: $$6 = 3 + \frac{C}{e^{2(1)}}.$$ Solve for C: $$C = 3(e^2 - 1).$$
06

Write the final solution for u(x)

Substitute the value of C back into the expression for \(u(x)\): $$u(x) = 3 + \frac{3(e^2-1)}{e^{2x}}.$$ This is the explicit solution for the function \(u(x)\) that satisfies the given ODE and initial condition.

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Most popular questions from this chapter

An object in free fall may be modeled by assuming that the only forces at work are the gravitational force and air resistance. By Newton's Second Law of Motion (mass \(\times\) acceleration \(=\) the sum of the external forces), the velocity of the object satisfies the differential equation $$\underbrace {m}_{\text {mass}}\quad \cdot \underbrace{v^{\prime}(t)}_{\text {acceleration }}=\underbrace {m g+f(v)}_{\text {external forces}}$$ where \(f\) is a function that models the air resistance (assuming the positive direction is downward). One common assumption (often used for motion in air) is that \(f(v)=-k v^{2},\) where \(k>0\) is a drag coefficient. a. Show that the equation can be written in the form \(v^{\prime}(t)=g-a v^{2},\) where \(a=k / m\) b. For what (positive) value of \(v\) is \(v^{\prime}(t)=0 ?\) (This equilibrium solution is called the terminal velocity.) c. Find the solution of this separable equation assuming \(v(0)=0\) and \(0 < v^{2} < g / a\) d. Graph the solution found in part (c) with \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) \(m=1,\) and \(k=0.1,\) and verify that the terminal velocity agrees with the value found in part (b).

The following models were discussed in Section 1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation \(m^{\prime}(t)+k m(t)=I,\) where \(m(t)\) is the mass of the drug in the blood at time \(t \geq 0, k\) is a constant that describes the rate at which the drug is absorbed, and \(I\) is the infusion rate. Let \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\). a. Draw the direction field, for \(0 \leq t \leq 100,0 \leq y \leq 600\). b. What is the equilibrium solution? c. For what initial values \(m(0)=A\) are solutions increasing? Decreasing?

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 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 if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y^{\prime}(x)=\sqrt{\frac{x+1}{y+4}}, y(3)=5$$

Explain how the growth rate function determines the solution of a population model.
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