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Chapter 2: Problem 10

For \(y^{\prime}+\frac{2}{x} y=\frac{3}{x}\) an integrating factor is \(e^{\int(2 / x) d x}=x^{2}\) so that \(\frac{d}{d x}\left[x^{2} y\right]=3 x\) and \(y=\frac{3}{2}+c x^{-2}\) for \(0< x<\infty\).

Short Answer

Expert verified
The solution is \( y = \frac{3}{2} + Cx^{-2} \).

Step by step solution

01

Identify the form of the differential equation

The given differential equation is first-order linear and can be written as: \( y' + P(x)y = Q(x) \), where \( P(x) = \frac{2}{x} \) and \( Q(x) = \frac{3}{x} \).
02

Determine the integrating factor

The integrating factor \( \mu(x) \) is determined by the formula \( e^{\int P(x) \, dx} \). Here \( P(x) = \frac{2}{x} \), so we calculate the integral: \[ \mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln|x|} = x^2. \]
03

Apply the integrating factor

Multiply the entire differential equation by the integrating factor \( x^2 \): \[ x^2 y' + x^2 \cdot \frac{2}{x}y = x^2 \cdot \frac{3}{x}. \] This simplifies to \[ \frac{d}{dx}[x^2 y] = 3x. \]
04

Integrate both sides

Integrate the equation from the previous step: \[ \int \frac{d}{dx}[x^2 y] \, dx = \int 3x \, dx. \] The left side simplifies to \( x^2 y \), and the right side integrates to \( \frac{3}{2}x^2 + C \). Hence, \( x^2 y = \frac{3}{2}x^2 + C \).
05

Solve for \( y \)

Divide both sides by \( x^2 \) to solve for \( y \): \[ y = \frac{3}{2} + \frac{C}{x^2}. \] Here, \( C \) is an arbitrary constant determined by initial conditions, though none are given in the problem.

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Key Concepts

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

Integrating Factor
The concept of an integrating factor is a crucial tool used to solve first-order linear differential equations. In such differential equations, the aim is to transform the equation in a way that allows easy integration. This is where the integrating factor comes into play.

For a differential equation of the form \( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by the formula:
  • \( \mu(x) = e^{\int P(x) \, dx} \)
The integrating factor essentially "modifies" the original equation so that the left-hand side becomes the derivative of a product of functions. This transformation is critical because it converts our equation into an easily integrable form.

In our given problem, the integrating factor was \( x^2 \), because the function \( P(x) = \frac{2}{x} \) integrates to \( 2 \ln |x| \), making \( \mu(x) = e^{2 \ln |x|} = x^2 \). This factor, when multiplied through the entire equation, aligns the equation nicely for further steps.
Differential Equations
Differential equations involve derivatives and are used to describe various phenomena that change over time or space. First-order linear differential equations, like the one in our example, involve only the first derivative of the unknown function. These equations typically have the structure:
  • \( y' + P(x)y = Q(x) \)
Where:
  • \( y' \) is the first derivative of \( y \) with respect to \( x \).
  • \( P(x) \) and \( Q(x) \) are functions of \( x \).
Understanding this structure is key to applying the correct method to solve them. They are prevalent in modeling situations in physics, engineering, biology, and economics, where rates of change are integral to the system dynamics.

For example, when faced with the task of solving the differential equation \( y' + \frac{2}{x} y = \frac{3}{x} \), we follow a systematic process that ultimately allows us to express \( y \) in terms of \( x \) by using specific solution techniques, such as integrating factors.
Solution Techniques
Solving first-order linear differential equations often involves a systematic procedure that starts with identifying the form of the equation, determining an integrating factor, and then applying it. Let鈥檚 break this process down:
  • Identify the Equation Form: Recognize that the equation is in the form \( y' + P(x)y = Q(x) \). This is crucial for selecting the correct method.
  • Find the Integrating Factor: Calculate \( \mu(x) = e^{\int P(x) \, dx} \). Here, it's \( x^2 \).
  • Multiply Through: Apply the integrating factor across the entire equation to transform it into a form where the left side is a product's derivative: \( \frac{d}{dx}[x^2y] = 3x \).
  • Integrate: Both sides are integrated with respect to \( x \), yielding a new equation where the left side integrates directly to a simple expression.
  • Solve for \( y \): The obtained expression is then solved for \( y \), often resulting in an equation that includes an arbitrary constant \( C \), accounting for initial conditions.
In our specific problem, after applying these steps, we find that \( y = \frac{3}{2} + \frac{C}{x^2} \), highlighting how systematic approaches allow us to peel back the complexity of differential equations and find solutions efficiently.

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Most popular questions from this chapter

(a) An implicit solution of the differential equation \((2 y+2) d y-\left(4 x^{3}+6 x\right) d x=0\) is \\[y^{2}+2 y-x^{4}-3 x^{2}+c=0\\]. The condition \(y(0)=-3\) implies that \(c=-3 .\) Therefore \(y^{2}+2 y-x^{4}-3 x^{2}-3=0\). (b) Using the quadratic formula we can solve for \(y\) in terms of \(x\): \\[y=\frac{-2 \pm \sqrt{4+4\left(x^{4}+3 x^{2}+3\right)}}{2}\\]. The explicit solution that satisfies the initial condition is then \\[y=-1-\sqrt{x^{4}+3 x^{3}+4}\\]. (c) From the graph of the function \(f(x)=x^{4}+3 x^{3}+4\) below we see that \(f(x) \leq 0\) on the approximate interval \(-2.8 \leq x \leq-1.3 .\) Thus the approximate domain of the function $$y=-1-\sqrt{x^{4}+3 x^{3}+4}=-1-\sqrt{f(x)}$$ is \(x \leq-2.8\) or \(x \geq-1.3 .\) The graph of this function is shown below. (d) Using the root finding capabilities of a CAS, the zeros of \(f\) are found to be -2.82202 and \(-1.3409 .\) The domain of definition of the solution \(y(x)\) is then \(x>-1.3409 .\) The equality has been removed since the derivative \(d y / d x\) does not exist at the points where \(f(x)=0 .\) The graph of the solution \(y=\phi(x)\) is given on the right.

(a) Writing the equation in the form \((x-\sqrt{x^{2}+y^{2}}) d x+y d y=0\) we identify \(M=x-\sqrt{x^{2}+y^{2}}\) and \(N=y\) since \(M\) and \(N\) are both homogeneous functions of degree 1 we use the substitution \(y=u x .\) It follows that \\[ \begin{aligned} (x-\sqrt{x^{2}+u^{2} x^{2}}) d x+u x(u d x+x d u) &=0 \\ x\left[1-\sqrt{1+u^{2}}+u^{2}\right] d x+x^{2} u d u &=0 \\ -\frac{u d u}{1+u^{2}-\sqrt{1+u^{2}}} &=\frac{d x}{x} \\ \frac{u d u}{\sqrt{1+u^{2}}(1-\sqrt{1+u^{2}})} &=\frac{d x}{x} \end{aligned} \\] Letting \(w=1-\sqrt{1+u^{2}}\) we have \(d w=-u d u / \sqrt{1+u^{2}}\) so that \\[ \begin{aligned} -\ln |1-\sqrt{1+u^{2}}| &=\ln |x|+c \\ \frac{1}{1-\sqrt{1+u^{2}}} &=c_{1} x \\ 1-\sqrt{1+u^{2}} &=-\frac{c_{2}}{x} \\ 1+\frac{c_{2}}{x} &=\sqrt{1+\frac{y^{2}}{x^{2}}} \\ 1+\frac{2 c_{2}}{x}+\frac{c_{2}^{2}}{x^{2}} &=1+\frac{y^{2}}{x^{2}} \end{aligned} \\] Solving for \(y^{2}\) we have \\[ y^{2}=2 c_{2} x+c_{2}^{2}=4\left(\frac{c_{2}}{2}\right)\left(x+\frac{c_{2}}{2}\right) \\] which is a family of parabolas symmetric with respect to the \(x\) -axis with vertex at \(\left(-c_{2} / 2,0\right)\) and focus at the origin. (b) Let \(u=x^{2}+y^{2}\) so that \\[ \frac{d u}{d x}=2 x+2 y \frac{d y}{d x} \\] Then \\[ y \frac{d y}{d x}=\frac{1}{2} \frac{d u}{d x}-x \\] and the differential equation can be written in the form \\[ \frac{1}{2} \frac{d u}{d x}-x=-x+\sqrt{u} \text { or } \frac{1}{2} \frac{d u}{d x}=\sqrt{u} \\] Separating variables and integrating gives $$\begin{aligned} \frac{d u}{2 \sqrt{u}} &=d x \\ \sqrt{u} &=x+c \\ u &=x^{2}+2 c x+c^{2} \\ x^{2}+y^{2} &=x^{2}+2 c x+c^{2} \\ y^{2} &=2 c x+c^{2} \end{aligned}$$

Write the differential equation as \(d P / d t-a P=-b P^{2}\) and let \(u=P^{-1}\) or \(P=u^{-1}\). Then $$\frac{d p}{d t}=-u^{-2} \frac{d u}{d t},$$ and substituting into the differential equation, we have $$-u^{-2} \frac{d u}{d t}-a u^{-1}=-b u^{-2} \quad \text { or } \quad \frac{d u}{d t}+a u=b.$$ The latter differential equation is linear with integrating factor \(e^{\int a d t}=e^{a t},\) so $$\frac{d}{d t}\left[e^{a t} u\right]=b e^{a t}$$ and $$\begin{aligned} e^{a t} u &=\frac{b}{a} e^{a t}+c \\ e^{a t} P^{-1} &=\frac{b}{a} e^{a t}+c \\ P^{-1} &=\frac{b}{a}+c e^{-a t} \\ P &=\frac{1}{b / a+c e^{-a t}}=\frac{a}{b+c_{1} e^{-a t}}. \end{aligned}$$

Assume \(R d q / d t+(1 / C) q=E(t), R=200, C=10^{-4},\) and \(E(t)=100\) so that \(q=1 / 100+c e^{-50 t} .\) If \(q(0)=0\) then \(c=-1 / 100\) and \(i=\frac{1}{2} e^{-50 t}\).

Assume that \(A=A_{0} e^{k t}\) and \(k=-0.00012378\). If \(A(t)=0.145 A_{0}\) then \(t \approx 15,600\) years.
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