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Chapter 8: Problem 1

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x^{2} y^{\prime}-x y=1 / x$$

Short Answer

Expert verified
The differential equation is linear first-order, and the solution is \( y = -\frac{1}{3x^{2}} + Cx \).

Step by step solution

01

- Identify the Type of Differential Equation

Identify the type of the given differential equation: \[x^{2} y^{\text{'}} - x y = \frac{1}{x}\]Rewriting it as: \[y^{\text{'}} - \frac{y}{x} = \frac{1}{x^{3}}\]This is a linear first-order differential equation.
02

- Standard Form

Rewrite the differential equation in standard form:\[y^{\text{'}} + P(x)y = Q(x)\]Here, \( P(x) = -\frac{1}{x} \) and \( Q(x) = \frac{1}{x^{3}} \). The equation is then:\[y^{\text{'}} - \frac{y}{x} = \frac{1}{x^{3}}\].
03

- Find the Integrating Factor

Find the integrating factor \( \mu(x) \) using the formula:\[ \mu(x) = e^{\int P(x) dx} \]Given \( P(x) = -\frac{1}{x} \), we have:\[ \mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x} \].
04

- Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \mu(x) = \frac{1}{x} \):\[ \frac{1}{x} y^{\text{'}} - \frac{y}{x^{2}} = \frac{1}{x^{4}} \].This simplifies to:\[ (\frac{y}{x})^{\text{'}} = \frac{1}{x^{4}} \].
05

- Integrate Both Sides

Integrate both sides with respect to \(x\):\[ \int (\frac{y}{x})^{\text{'}} dx = \int \frac{1}{x^{4}} dx \].We get:\[ \frac{y}{x} = \int x^{-4} dx = \frac{x^{-3}}{-3} + C = -\frac{1}{3x^{3}} + C \].
06

- Solve for y

Solve for \(y\) by multiplying both sides by \( x \):\[ y = x \left(-\frac{1}{3x^{3}} + C\right) = -\frac{1}{3x^{2}} + Cx \].

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Key Concepts

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

Integrating Factor
An integrating factor is a function we use to simplify the solution of a linear first-order differential equation. By multiplying the entire differential equation by this factor, we can turn the left-hand side of the equation into an exact derivative of a product of functions, making it much easier to integrate.

To find the integrating factor, use the formula:
\[ \mu(x) = e^{\textstyle{\int P(x) \ dx}} \]
\( P(x) \) is derived from the standard form of the linear differential equation. In our example, we had:
\[ P(x) = -\frac{1}{x} \]
which leads to:
\[ \mu(x) = e^{\textstyle{\int -\frac{1}{x} \ dx}} = e^{\textstyle{-\ln|x|}} = \frac{1}{x} \]
This integrating factor transforms the equation into a form where it can be easily integrated and solved.
Differential Equation Types
Differential equations come in different types, suitable for various methods of solution. Some common ones include:
  • Separable Differential Equations: Can be written as \( f(y)dy = g(x)dx \).
  • Linear First-Order: Of the form \( y' + P(x)y = Q(x) \).
  • Linear Second-Order: Typically looks like \( y'' + P(x)y' + Q(x)y = R(x) \).

In the given problem, we identified the equation:
\[ x^{2} y^{\text{'}} - x y = \frac{1}{x} \]
as a linear first-order differential equation. It's recognized by the form \( y' + P(x)y = Q(x) \), characterized by its first-order derivative and linearity in both \( y \) and \( y' \), making it amenable to methods like using integrating factors.
Solving Differential Equations
Solving differential equations involves several steps tailored to the type of equation. For a linear first-order differential equation like the one in the exercise, the steps are:
  • Identify the Equation Type: Determine if the differential equation is linear first-order, separable, etc.
  • Rewrite in Standard Form: Express it in the form \( y' + P(x)y = Q(x) \).
  • Find the Integrating Factor: Compute\( \mu(x) = e^{\textstyle{\int P(x) \ dx}} \).
  • Multiply Through by the Integrating Factor: This simplifies it to the form \( d/dx (\mu(x)y) = \mu(x)Q(x) \).
  • Integrate Both Sides: Find the solution by integrating both sides with respect to \( x \).
  • Solve for \( y \): Solve the integral to find \( y \) explicitly.

In the exercise:
\[ \int (\frac{y}{x})^{\text{'}} dx = \int \frac{1}{x^{4}} dx \]
This solves to:
\[ y = x (\frac{x^{-3}}{-3} + C) = -\frac{1}{3x^{2}} + Cx \]
Each step builds on the previous one, leading to the solution of the differential equation.

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Most popular questions from this chapter

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}-4 y=4 e^{2 t}, \quad y_{0}=0, y_{0}^{\prime}=1$$

Compare the temperatures of your cup of coffee at time \(t\) (a) if you add cream and let the mixture cool; (b) if you let the coffee and cream sit on the table and mix them at time \(t\). Hints: Assume Newton's law of cooling (Problem 2.27) for both coffee and cream (where it is a law of heating). Combine \(n^{\prime}\) units of cream initially at temperature \(T_{0}^{\prime}\) with \(n\) units of coffee initially at temperature \(T_{0},\) and find the temperature at time \(t\) in (a) and in (b) assuming that the air temperature remains a constant \(T_{a},\) and that the proportionality constant in the law of cooling is the same for both coffee and cream.

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+9 y=\cos 3 t, \quad y_{0}=0, y_{0}^{\prime}=6$$

(a) Show that $$\begin{aligned} D\left(e^{a x} y\right) &=e^{a x}(D+a) y \\ D^{2}\left(e^{a x} y\right) &=e^{a x}(D+a)^{2} y \end{aligned}$$ and so on; that is, for any positive integral \(n\) $$D^{n}\left(e^{a x} y\right)=e^{a x}(D+a)^{n} y$$ Thus show that if \(L(D)\) is any polynomial in the operator \(D,\) then $$L(D)\left(e^{a x} y\right)=e^{a x} L(D+a) y$$ This is called the exponential shift. (b) Use (a) to show that $$\begin{aligned} (D-1)^{3}\left(e^{x} y\right) &=e^{x} D^{3} y \\ \left(D^{2}+D-6\right)\left(e^{-3 x} y\right) &=e^{-3 x}\left(D^{2}-5 D\right) y \end{aligned}$$ (c) Replace \(D\) by \(D-a,\) to obtain $$e^{a x} P(D) y=P(D-a) e^{a x} y$$ This is called the inverse exponential shift. (d) Using (c), we can change a differential equation whose right-hand side is an exponential times a polynomial, to one whose right-hand side is just a polynomial. For example, consider \(\left(D^{2}-D-6\right) y=10 x e^{3 x} ;\) multiplying both sides by \(e^{-3 x}\) and using \((\mathrm{c}),\) we get $$\begin{aligned} e^{-3 x}\left(D^{2}-D-6\right) y &=\left[(D+3)^{2}-(D+3)-6\right] y e^{-3 x} \\\ &=\left(D^{2}+5 D\right) y e^{-3 x}=10 x \end{aligned}$$ Show that a solution of \(\left(D^{2}+5 D\right) u=10 x\) is \(u=x^{2}-\frac{2}{5} x ;\) then \(y e^{-3 x}=x^{2}-\frac{2}{5} x\) or \(y=e^{3 x}\left(x^{2}-\frac{2}{5} x\right) .\) Use this method to solve Problems 23 to 26

For each of the following differential equations, separate variables and find a solution containing one arbitrary constant. Then find the value of the constant to give a particular solution satisfying the given boundary condition. Computer plot a slope field and some of the solution curves. \(y^{\prime}-x y=x, \quad y=1\) when \(x=0\)
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