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Chapter 4: Problem 46

\(y^{\prime \prime}+3 y^{\prime}=18, y(0)=0, y^{\prime}(0)=3\)

Short Answer

Expert verified
The solution is \[ y = 6t + e^{-3t} - 1. \]

Step by step solution

01

Rewrite the differential equation

Given the second-order differential equation \[ y'' + 3y' = 18 \] Rewrite it as a first-order system by letting \[ y' = v \] and consequently \[ y'' = v' \].Thus, we have \[ v' + 3v = 18. \]
02

Solve the first-order differential equation for v

To solve \[ v' + 3v = 18, \] uniformly multiply both sides by the integrating factor \[ e^{3t}. \]This results in \[ \frac{d}{dt} \big( v e^{3t} \big) = 18 e^{3t}. \]Integrate both sides with respect to t: \[ v e^{3t} = \frac{18}{3} e^{3t} + C \] which simplifies to \[ v e^{3t} = 6 e^{3t} + C. \]Thus, the solution for v is \[ v = 6 + Ce^{-3t}. \]
03

Use initial condition to find C

Given \[ y'(0) = 3, \]substitute t = 0 into \[ v = 6 + Ce^{-3t}: \] \[ 3 = 6 + C \]Solve for C: \[ C = -3. \]So, \[ v = 6 - 3e^{-3t}. \]
04

Integrate v to find y

Since \[ y' = v = 6 - 3e^{-3t}, \]Integrate \[ y' = 6 - 3e^{-3t} \]with respect to t to find y: \[ y = \frac{6t}{1!} - \frac{3}{-3} e^{-3t} + K \]This simplifies to \[ y = 6t + e^{-3t} + K. \]
05

Use initial condition to find K

Given \[ y(0) = 0, \]substitute t = 0 into \[ y = 6t + e^{-3t} + K: \] \[ 0 = 6(0) + e^{0} + K \]Solve for K: \[ K = -1. \]Thus, the solution for y is \[ y = 6t + e^{-3t} - 1. \]

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Key Concepts

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

first-order system
To understand the process of converting a second-order differential equation into a first-order system, let's begin with a given equation. For example, consider the second-order differential equation:

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Most popular questions from this chapter

(a) Show that the transformed equation for the Cauchy-Euler equation \(a x^{2} y^{\prime \prime}+b x y^{\prime}+\) \(c y=0\) is \(a d^{2} y / d t^{2}+(b-a) d y / d t+c y=\) 0 . (b) Show that if the characteristic equation for a transformed Cauchy-Euler equation has a repeated root, then this root is \(r=-(b-a) /(2 a)\). (c) Use reduction of order to show that a second solution to this equation is \(x^{r} \ln x\). (d) Calculate the Wronskian of \(x^{r}\) and \(x^{r} \ln x\) to show that these two solutions are linearly independent.

Let \(x=-e^{t}\). (a) Show that \(\frac{d y}{d x}=\frac{d y}{d t} \frac{d t}{d x}=\) \(\frac{1}{x} \frac{d y}{d t}\) and \(\frac{d^{2} y}{d x^{2}}=\frac{1}{x^{2}}\left(\frac{d^{2} y}{d t^{2}}-\frac{d y}{d t}\right)\). (b) Show that the differential equation \(a x^{2} y^{\prime \prime}+b x y^{\prime}+\) \(c y=f(x)\) is transformed into \(a d^{2} y / d t^{2}+\) \((b-a) d y / d t+c y=f\left(-e^{t}\right)\).

\(x^{3} y^{\prime \prime \prime}-6 x^{2} y^{\prime \prime}+17 x y^{\prime}-17 y=0, y(1)=-2\), \(y^{\prime}(1)=0, y^{\prime \prime}(1)=0\)

\(2 x^{2} y^{\prime \prime}+3 x y^{\prime}-y=x^{-2}, y(1)=0, y^{\prime}(1)=2\)

Solve each of the following initial value problems. Verify that your result satisfies the initial conditions by graphing it on an appropriate interval. (a) \(y^{\prime \prime \prime}+3 y^{\prime \prime}+2 y^{\prime}+6 y=0, y(0)=0, y^{\prime}(0)=\) \(1, y^{\prime \prime}(0)=-1\) (b) \(y^{(4)}-8 y^{\prime \prime \prime}+30 y^{\prime \prime}-56 y^{\prime}+49 y=0\), \(y(0)=1, y^{\prime}(0)=2, y^{\prime \prime}(0)=-1, y^{\prime \prime \prime}(0)=\) \(-1\) (c) \(0.31 y^{\prime \prime \prime}+11.2 y^{\prime \prime}-9.8 y^{\prime}+5.3 y=0\), \(y(0)=-1, y^{\prime}(0)=-1, y^{\prime \prime}(0)=0\)
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