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Chapter 8: Problem 11

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$(y+2 x) d x-x d y=0$$

Short Answer

Expert verified
The given equation is separable. The solution is y+2x=m x ln y=-2

Step by step solution

01

Identify the type of differential equation

The given differential equation is d(x+y)=0 Rewriting it gives us:dy/dx = f(x)dy=-x/(2x) which identifies it as a separable differential equation since it can be written as dy = g(x)dx, and the variables can be separated.
02

Separate the variables

Separate the variables by dividing both sides by (y+2x).dx/(y+2x)d(y+2x=1)ydx/dxy-(2in)dx/dx=dxd(x)+2dx=0
03

Integrate both sides

Integrate both sides of the equation: pm the function gives. $$g()$ m+Int dx = c( x*x.0-d8pm*dx)
04

Solve for the constant.

Rewrite the integrated terms to solve for the constant: y+2x-c8dx = ln (c+p) +c8.
05

Simplify the solution

Interestingly, it can be written asy y=-2xln(m*x)

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Key Concepts

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

First-Order Differential Equations
First-order differential equations involve the first derivative of a function but no higher-order derivatives.
In mathematical terms, they can be written in the form:
\[ \frac{dy}{dx} = f(x, y) \]
These equations describe how a function changes with respect to an independent variable.
Common types of first-order differential equations include separable, linear, and exact.
In the given exercise, the differential equation can be identified as separable because the variables can be separated on opposite sides of the equation.
Recognizing the type of differential equation helps in choosing the right method to solve it.
Integration
Integration is a fundamental concept in calculus used to find the antiderivative or the area under a curve.
In the context of solving differential equations, integration is applied after separating variables.
Once you have separated the variables, you integrate both sides with respect to their respective variables.
For instance, if we have:
\[ \frac{dy}{dx} = g(x)h(y) \]
We separate the variables as:
\[ \frac{1}{h(y)} dy = g(x) dx \]
Then, we integrate both sides:
\[ \begin{align*} \text{Integrate:} \ \frac{1}{h(y)} dy &= \text{Integral of } \frac{1}{h(y)} \text{ with respect to } y \ g(x) dx &= \text{Integral of } g(x) \text{ with respect to } x \end{align*} \]
This process gives us the solution that usually involves a constant of integration.
Clearly separating and then integrating is crucial in solving these problems.
Solving Differential Equations
Solving a differential equation involves finding a function that satisfies the given equation.
The steps usually include:
  • Identifying the type of differential equation.
  • Separating variables (if possible).
  • Integrating both sides.
  • Solving for the constant of integration.
  • Simplifying the solution.

In the given exercise:
  • The equation was identified as a separable differential equation.
  • Variables were separated: \[ (y+2x) dx - x dy = 0 \]
  • The equation was rearranged and integrated: \[ \frac{dy}{dx} = -\frac{x}{y+2x} \]
  • Both sides were integrated and simplified to find the general solution: \[ y + 2x = C \]
Solving differential equations often involves these systematic steps.
Each step needs to be carefully executed for accurate results.
Always verify your final solution by plugging it back into the original equation.

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Most popular questions from this chapter

Suppose that the rate at which you work on a hot day is inversely proportional to the excess temperature above \(75^{\circ} .\) One day the temperature is rising steadily, and you start studying at 2 p.m. You cover 20 pages the first hour and 10 pages the second hour. At what time was the temperature \(75^{\circ} ?\)

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top and bottom of the block are parallel planes which remain horizontal during the oscillations and that the sides of the block are vertical. Show that the period of the motion (neglecting friction) is \(2 \pi \sqrt{h / g}\) where \(h\) is the vertical height of the part of the block under water when it is floating at rest. Hint: Recall that the buoyant force is equal to the weight of displaced water.

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

A mass \(m\) falls under gravity (force \(m g\) ) through a liquid whose viscosity is decreasing so that the retarding force is \(-2 m v /(1+t),\) where \(v\) is the speed of \(m .\) If the mass starts from rest, find its speed, its acceleration, and how far it has fallen (in terms of \(g\) ) when \(t=1\).

(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
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