/*! 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 21 From \(y^{\prime}-\frac{2}{x} y=... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 21

From \(y^{\prime}-\frac{2}{x} y=\frac{3}{x^{2}} y^{4}\) and \(w=y^{-3}\) we obtain \(\frac{d w}{d x}+\frac{6}{x} w=-\frac{9}{x^{2}} .\) An integrating factor is \(x^{6}\) so that \(x^{6} w=-\frac{9}{5} x^{5}+c\) or \(y^{-3}=-\frac{9}{5} x^{-1}+c x^{-6} .\) If \(y(1)=\frac{1}{2}\) then \(c=\frac{49}{5}\) and \(y^{-3}=-\frac{9}{5} x^{-1}+\frac{49}{5} x^{-6}.\)

Short Answer

Expert verified
The solution is \( y^{-3} = \frac{-9}{5} x^{-1} + \frac{49}{5} x^{-6} \).

Step by step solution

01

Understand the Given Differential Equation

We start with the first-order nonlinear differential equation: \( y^{\prime}-\frac{2}{x} y=\frac{3}{x^{2}} y^{4} \) and the substitution \( w = y^{-3} \). This substitution will simplify the equation.
02

Substitute and Transform the Equation

By substituting \( w = y^{-3} \), we find \( \frac{dw}{dx} = -3y^{-4} \frac{dy}{dx} \). Using this, the differential equation transforms into \( \frac{d w}{d x} + \frac{6}{x} w = -\frac{9}{x^{2}}. \)
03

Identify and Apply the Integrating Factor

The given integrating factor is \( x^{6} \). When applied to the equation \( \frac{d w}{d x} + \frac{6}{x} w = -\frac{9}{x^{2}} \), it becomes an exact equation: \( \frac{d}{dx}(x^6 w) = -\frac{9}{x^2} \cdot x^6. \)
04

Integrate Both Sides

Integrate the equation \( \frac{d}{dx}(x^6 w) = -9x^4 \) with respect to \( x \). This gives \( x^6 w = -\frac{9}{5} x^5 + C, \) where \( C \) is the constant of integration.
05

Solve for \( w \) in Terms of \( x \)

Rewriting the expression gives \( w = \frac{-9}{5x} + \frac{C}{x^6}. \) Since \( w = y^{-3}, \) the expression becomes \( y^{-3} = \frac{-9}{5} x^{-1} + C x^{-6}. \)
06

Apply the Initial Condition to Find \( C \)

The initial condition \( y(1) = \frac{1}{2} \) means \( y^{-3} = 2^3 = 8. \) Substituting these into the equation gives: \( 8 = \frac{-9}{5}(1) + C(1)^6. \) Solving for \( C \) provides \( C = \frac{49}{5}. \)
07

State the Final Solution

The expression for \( y^{-3} \) in terms of \( x \) with the constant \( C = \frac{49}{5} \) is: \( y^{-3} = \frac{-9}{5} x^{-1} + \frac{49}{5} x^{-6}. \)

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Key Concepts

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

Substitution Method
The substitution method is a powerful technique in dealing with nonlinear differential equations. By introducing a new variable, the equation can be transformed into a simpler form.
The original problem contains a complex nonlinear term with a power of four, represented by the equation: \( y^{\prime}-\frac{2}{x} y=\frac{3}{x^{2}} y^{4} \).
With the substitution \( w = y^{-3} \), the equation becomes more manageable.
This transformation alters the equation
and allows us to work with a linear form: \( \frac{dw}{dx} + \frac{6}{x} w = -\frac{9}{x^{2}} \).
  • Think of substitution as changing the lens through which you see the problem.
  • It often simplifies solving or even makes an impossible-seeming equation solvable!
  • In this case, the power of substitution lies in converting a nonlinear equation to a linear one.
Integrating Factor
To solve the transformed linear differential equation, an integrating factor is applied. An integrating factor helps to rewrite a linear first-order differential equation into an exact equation
and allows us to integrate both sides easily. In our case, the integrating factor is \( x^{6} \).
The equation \( \frac{dw}{dx} + \frac{6}{x} w = -\frac{9}{x^{2}} \) is multiplied by the integrating factor:
\[ (x^{6})\left(\frac{dw}{dx} \right) + (x^{6})\left(\frac{6}{x} w\right) = (x^{6})\left(-\frac{9}{x^{2}}\right). \]
This converts the left-hand side into the derivative of a product, \( \frac{d}{dx}(x^6 w) = -9x^4 \).
  • The integrating factor effectively balances the equation.
  • It essentially "unlocks" the differentials for smooth integration.
  • Think of it as a key for exactness in differential equations.
Initial Condition
Initial conditions are crucial in differential equations as they help us determine the specific solution that satisfies the problem's constraints.
For our problem, the initial condition is given as \( y(1) = \frac{1}{2} \).
This tells us exactly what the solution must satisfy at this specific point. By substituting this into our problem, we derive \( y^{-3} = 2^3 = 8 \).
Next, plug into the solved equation, \( 8 = \frac{-9}{5}(1) + C(1)^6 \), to find the constant \( C \), resulting in \( C = \frac{49}{5} \).
Without this critical piece, our solution could represent any number of possibilities.
  • Initial conditions pinpoint your solution out of many possible ones.
  • They often represent real-world constraints or measurements.
  • Never ignore initial conditions鈥攖hey give your solution meaning, connecting it to reality!
Exact Equation
An exact equation is one where differentiation and integration are seamlessly intertwined, allowing us to recover a smooth solution.
With an integrating factor \( x^{6} \), our equation \( \frac{dw}{dx} + \frac{6}{x} w = -\frac{9}{x^{2}} \) becomes exact.
This means the left-hand side is the derivative of a single expression: \( \frac{d}{dx}(x^6 w) \).
Integrating both sides is straightforward: \[ \int \frac{d}{dx}(x^6 w) \ dx = \int -9x^4 \ dx. \]
This results in \( x^6 w = -\frac{9}{5} x^5 + C \).
  • Exact equations are simple once the proper integrating factor is applied.
  • They provide a clear path from differentiation to integration.
  • Always check for exactness to simplify solving processes!

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Most popular questions from this chapter

(a) The initial temperature of the bath is \(T_{m}(0)=60^{\circ},\) so in the short term the temperature of the chemical, which starts at \(80^{\circ},\) should decrease or cool. Over time, the temperature of the bath will increase toward \(100^{\circ}\) since \(e^{-0.1 t}\) decreases from 1 toward 0 as \(t\) increases from \(0 .\) Thus, in the long term, the temperature of the chemical should increase or warm toward \(100^{\circ}\). (b) Adapting the model for Newton's law of cooling, we have $$\frac{d T}{d t}=-0.1\left(T-100+40 e^{-0.1 t}\right), \quad T(0)=80$$. Writing the differential equation in the form $$\frac{d T}{d t}+0.1 T=10-4 e^{-0.1 t}$$ we see that it is linear with integrating factor \(e^{\int 0.1 d t}=e^{0.1 t}\) Thus $$\begin{aligned} \frac{d}{d t}\left[e^{0.1 t} T\right] &=10 e^{0.1 t}-4 \\ e^{0.1 t} T &=100 e^{0.1 t}-4 t+c \end{aligned}$$ and $$T(t)=100-4 t e^{-0.1 t}+c e^{-0.1 t}$$ Now \(T(0)=80\) so \(100+c=80, c=-20\) and $$T(t)=100-4 t e^{-0.1 t}-20 e^{-0.1 t}=100-(4 t+20) e^{-0.1 t}$$ The thinner curve verifies the prediction of cooling followed by warming toward \(100^{\circ} .\) The wider curve shows the temperature \(T_{m}\) of the liquid bath.

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

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.

(a) Solving \(v_{t}=\sqrt{m g / k}\) for \(k\) we obtain \(k=m g / v_{t}^{2} .\) The differential equation then becomes \\[ m \frac{d v}{d t}=m g-\frac{m g}{v_{t}^{2}} v^{2} \\] or \(\frac{d v}{d t}=g\left(1-\frac{1}{v_{t}^{2}} v^{2}\right)\) Separating variables and integrating gives \\[ v_{t} \tanh ^{-1} \frac{v}{v_{t}}=g t+c_{1} \\] The initial condition \(v(0)=0\) implies \(c_{1}=0,\) so \\[ v(t)=v_{t} \tanh \frac{g t}{v_{t}} \\] We find the distance by integrating: \\[ s(t)=\int v_{t} \tanh \frac{g t}{v_{t}} d t=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)+c_{2} \\] The initial condition \(s(0)=0\) implies \(c_{2}=0,\) so \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right) \\] In 25 seconds she has fallen \(20,000-14,800=5,200\) feet. Using a CAS to solve \\[ 5200=\left(v_{t}^{2} / 32\right) \ln \left(\cosh \frac{32(25)}{v_{t}}\right) \\] for \(v_{t}\) gives \(v_{t} \approx 271.711 \mathrm{ft} / \mathrm{s} .\) Then \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)=2307.08 \ln (\cosh 0.117772 t) \\] (b) \(\operatorname{At} t=15, s(15)=2,542.94 \mathrm{ft}\) and \(v(15)=s^{\prime}(15)=256.287 \mathrm{ft} / \mathrm{sec}.\)

From \(y e^{y} d y=\left(e^{-x}+e^{-3 x}\right) d x\) we obtain \(y e^{y}-e^{y}+e^{-x}+\frac{1}{3} e^{-3 x}=c\).
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