/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 From \(y^{\prime}+\frac{1}{x} y=... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 15

From \(y^{\prime}+\frac{1}{x} y=\frac{1}{x} y^{-2}\) and \(w=y^{3}\) we obtain \(\frac{d w}{d x}+\frac{3}{x} w=\frac{3}{x} .\) An integrating factor is \(x^{3}\) so that \(x^{3} w=x^{3}+c\) or \(y^{3}=1+c x^{-3}\).

Short Answer

Expert verified
The solution is \( y^3 = 1 + C x^{-3} \).

Step by step solution

01

Identifying the Equation

We are given the differential equation \( y' + \frac{1}{x} y = \frac{1}{x} y^{-2} \) and the transformation \( w = y^3 \). When we apply the substitution, we obtain a new differential equation for \( w \): \( \frac{d w}{dx} + \frac{3}{x} w = \frac{3}{x} \). This is a first-order linear differential equation.
02

Finding the Integrating Factor

To solve the linear differential equation, we first need an integrating factor. The integrating factor for an equation of the form \( \frac{d w}{dx} + P(x) w = Q(x) \) is given by \( e^{\int P(x) \, dx} \). Here, \( P(x) = \frac{3}{x} \), so the integrating factor is \( e^{\int \frac{3}{x} \, dx} = e^{3 \ln|x|} = |x|^3 \). Since \( x^3 \) is the provided integrating factor, we can use it.
03

Applying the Integrating Factor

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

Integrating Both Sides

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx} (x^3 w) \, dx = \int 3x^2 \, dx \). The left side simplifies to \( x^3 w \), and the right side integrates to \( x^3 + C \), giving the equation \( x^3 w = x^3 + C \).
05

Solving for Original Function

Solving for \( w \) we get \( w = 1 + \frac{C}{x^3} \). Since \( w = y^3 \), substitute back to get \( y^3 = 1 + \frac{C}{x^3} \). This is the solution expressed in terms of the original variable \( y \).

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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 vital in solving first-order linear differential equations. An integrating factor is essentially a function that, when multiplied with a given differential equation, transforms it into an exact equation, which is easier to integrate. To find the integrating factor, consider a linear differential equation of the form \( \frac{dw}{dx} + P(x)w = Q(x) \).
  • Determine \( P(x) \), which is the coefficient of \( w \) in the equation.
  • The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} \).
For instance, in our exercise, \( P(x) = \frac{3}{x} \). Calculating the integrating factor involves:\[ \mu(x) = e^{\int \frac{3}{x} \, dx} = e^{3 \ln|x|} = |x|^3 \]This integrating factor, \( x^3 \), plays a crucial role in solving the equation as it allows us to write the left-hand side as the derivative of a product, facilitating straightforward integration.
First-Order Linear Differential Equations
First-order linear differential equations are equations that involve the first derivative of the unknown function, and they are linear in terms of that function and its first derivative. A typical first-order linear differential equation can be expressed as:\[ \frac{dy}{dx} + P(x)y = Q(x) \]Such equations are common in various fields, from physics to economics. The main goal is to find the function \( y(x) \) that satisfies the equation, given \( P(x) \) and \( Q(x) \). In the example from our exercise, after substituting \( w = y^3 \), we end up with the first-order linear differential equation:\[ \frac{dw}{dx} + \frac{3}{x} w = \frac{3}{x} \]To solve, we use the integrating factor method discussed previously. The integrating transforms the equation into one that can be directly integrated, which provides a path to finding the solution.
Variable Transformation
Variable transformation is a powerful technique used to simplify complex differential equations by substituting one variable with another. By using this method, we aim to convert an original problem into a form that is easier to handle.
  • The original equation \( y' + \frac{1}{x} y = \frac{1}{x}y^{-2} \) can be difficult to solve directly.
  • Instead, we substitute \( w = y^3 \), which transforms the equation into: \( \frac{dw}{dx} + \frac{3}{x} w = \frac{3}{x} \).
This transformation results in an equation that fits the form required for applying the integrating factor method. By changing the variables, we reduce the order or complexity, making integration possible. Once resolved for \( w \), we return to the original variables by reversing the substitution, obtaining a solution in terms of \( y \). This method is particularly useful when dealing with equations that are challenging to solve directly, giving us a clearer path to the solution.

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Most popular questions from this chapter

Assuming that the air resistance is proportional to velocity and the positive direction is downward with \(s(0)=0\) the model for the velocity is \(m d v / d t=m g-k v .\) Using separation of variables to solve this differential equation, we obtain \(v(t)=m g / k+c e^{-k t / m} .\) Then, using \(v(0)=0,\) we get \(v(t)=(m g / k)\left(1-e^{-k t / m}\right)\) Letting \(k=0.5, m=(125+35) / 32=5,\) and \(g=32,\) we have \(v(t)=320\left(1-e^{-0.1 t}\right) .\) Integrating we find \(s(t)=320 t+3200 e^{-0.1 t}+c_{1} . \quad\) Solving \(s(0)=0\) for \(c_{1}\) we find \(c_{1}=-3200,\) therefore \(s(t)=\) \(320 t+3200 e^{-0.1 t}-3200 .\) At \(t=15,\) when the parachute opens, \(v(15)=248.598\) and \(s(15)=2314.02\) At this time the value of \(k\) changes to \(k=10\) and the new initial velocity is \(v_{0}=248.598 .\) With the parachute open, the skydiver's velocity is \(v_{p}(t)=m g / k+c_{2} e^{-k t / m},\) where \(t\) is reset to 0 when the parachute opens. Letting \(m=5, g=32,\) and \(k=10,\) this gives \(v_{p}(t)=16+c_{2} e^{-2 t} .\) From \(v(0)=248.598\) we find \(c_{2}=232.598\) so \(v_{p}(t)=16+232.598 e^{-2 t} .\) Integrating, we get \(s_{p}(t)=16 t-116.299 e^{-2 t}+c_{3} .\) Solving \(s_{p}(0)=0\) for \(c_{3}\) we find \(c_{3}=116.299,\) so \(s_{p}(t)=16 t-116.299 e^{-2 t}+116.299 .\) Twenty seconds after leaving the plane is five seconds after the parachute opens. The skydiver's velocity at this time is \(v_{p}(5)=16.0106 \mathrm{ft} / \mathrm{s}\) and she has fallen a total of \(s(15)+s_{p}(5)=2314.02+196.294=2510.31 \mathrm{ft} .\) Her terminal velocity is \(\lim _{t \rightarrow \infty} v_{p}(t)=16, \mathrm{s}\) she has very nearly reached her terminal velocity five seconds after the parachute opens. When the parachute opens, the distance to the ground is \(15,000-s(15)=15,000-2,314=12,686\) ft. Solving \(s_{p}(t)=12,686\) we get \(t=785.6 \mathrm{s}=13.1 \mathrm{min} .\) Thus, it will take her approximately 13.1 minutes to reach the ground after her parachute has opened and a total of \((785.6+15) / 60=13.34\) minutes after she exits the plane.

(a) Solving \(k_{1}(M-A)-k_{2} A=0\) for \(A\) we find the equilibrium solution \(A=k_{1} M /\left(k_{1}+k_{2}\right) .\) From the phase portrait we see that \(\lim _{t \rightarrow \infty} A(t)=k_{1} M /\left(k_{1}+k_{2}\right)\) since \(k_{2}>0,\) the material will never be completely memorized and the larger \(k_{2}\) is, the less the amount of material will be memorized over time. (b) Write the differential equation in the form \(d A / d t+\left(k_{1}+k_{2}\right) A=k_{1} M\) Then an integrating factor is \(e^{\left(k_{1}+k_{2}\right) t},\) and $$\begin{aligned} \frac{d}{d t}\left[e^{\left(k_{1}+k_{2}\right) t} A\right] &=k_{1} M e^{\left(k_{1}+k_{2}\right) t} \\ e^{\left(k_{1}+k_{2}\right) t} A &=\frac{k_{1} M}{k_{1}+k_{2}} e^{\left(k_{1}+k_{2}\right) t}+c \\ A &=\frac{k_{1} M}{k_{1}+k_{2}}+c e^{-\left(k_{1}+k_{2}\right) t} \end{aligned}$$ \(\operatorname{Using} A(0)=0\) we find \(c=-\frac{k_{1} M}{k_{1}+k_{2}}\) and \(A=\frac{k_{1} M}{k_{1}+k_{2}}\left(1-e^{-\left(k_{1}+k_{2}\right) t}\right) .\) As \(t \rightarrow \infty\) \(A \rightarrow \frac{k_{1} M}{k_{1}+k_{2}}\).

Differentiating \(\ln \left(x^{2}+10\right)+\csc y=c\) we get \\[\begin{array}{c}\frac{2 x}{x^{2}+10}-\csc y \cot y \frac{d y}{d x}=0 \\\\\frac{2 x}{x^{2}+10}-\frac{1}{\sin y} \cdot \frac{\cos y}{\sin y} \frac{d y}{d x}=0\end{array}\\] or \\[2 x \sin ^{2} y d x-\left(x^{2}+10\right) \cos y d y=0\\] Writing the differential equation in the form \(\frac{d y}{d x}=\frac{2 x \sin ^{2} y}{\left(x^{2}+10\right) \cos y}\) we see that singular solutions occur when \(\sin ^{2} y=0,\) or \(y=k \pi,\) where \(k\) is an integer.

From \(d A / d t=4-A / 50\) we obtain \(A=200+c e^{-t / 50} .\) If \(A(0)=30\) then \(c=-170\) and \(A=200-170 e^{-t / 50}\).

(a) The initial-value problem is \(d h / d t=-8 A_{h} \sqrt{h} / A_{w}, h(0)=H\) Separating variables and integrating we have \\[ \frac{d h}{\sqrt{h}}=-\frac{8 A_{h}}{A_{w}} d t \quad \text { and } \quad 2 \sqrt{h}=-\frac{8 A_{h}}{A_{w}} t+c \\] Using \(h(0)=H\) we find \(c=2 \sqrt{H},\) so the solution of the initial-value problem is \(\sqrt{h(t)}=\left(A_{w} \sqrt{H}-4 A_{h} t\right) / A_{w},\) where \(A_{w} \sqrt{H}-4 A_{h} t \geq 0 .\) Thus, \\[ h(t)=\left(A_{w} \sqrt{H}-4 A_{h} t\right)^{2} / A_{w}^{2} \quad \text { for } \quad 0 \leq t \leq A_{w} H / 4 A_{h} \\] (b) Identifying \(H=10, A_{w}=4 \pi,\) and \(A_{h}=\pi / 576\) we have \(h(t)=t^{2} / 331,776-(\sqrt{5 / 2} / 144) t+10 .\) Solving \(h(t)=0\) we see that the tank empties in \(576 \sqrt{10}\) seconds or 30.36 minutes.
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