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In Problems \(1-10\), solve the given differential equation by using an appropriate substitution. $$ (x-y) d x+x d y=0 $$

Short Answer

Expert verified
The implicit solution is \(x = \frac{y-x}{C}\), where \(C\) is a constant.

Step by step solution

01

Identify the substitution method

The given differential equation is \((x-y) \, dx + x \, dy = 0\). This can be solved using the substitution method by recognizing it as a homogeneous equation. We substitute \(y = vx\) where \(v\) is a function of \(x\), making the equation homogeneous.
02

Differentiate the substitution

Differentiate the substitution \(y = vx\) with respect to \(x\), giving \(dy = v \, dx + x \, dv\). Substitute this into the original differential equation.
03

Substitute into the original equation

Substituting \(y = vx\) and \(dy = v \, dx + x \, dv\) into the original equation gives \((x - vx) \, dx + x(v \, dx + x \, dv) = 0\).
04

Simplify the equation

Factor the equation from Step 3: \(x(1 - v) \, dx + xv \, dx + x^2 \, dv = 0\). Simplifying gives \((1-v) \, dx + x \, dv = 0\).
05

Rearrange and separate variables

Rearrange the equation for separation of variables: \((1-v) \, dx + x \, dv = 0\) becomes \(\frac{dx}{x} = \frac{dv}{v-1}\).
06

Integrate both sides

Integrate both sides: \(\int \frac{dx}{x} = \int \frac{dv}{v-1}\). This gives \(\ln|x| = \ln|v-1| + C\) where \(C\) is a constant of integration.
07

Solve for the integration constant

Solve the integration equation \(\ln|x| = \ln|v-1| + C\) by exponentiating both sides resulting in \(|x| = e^C |v-1|\). Let \(e^C = k\), where \(k\) is an arbitrary constant. Therefore, \(|x| = k|v-1|\).
08

Substitute back for y

Since \(v = \frac{y}{x}\), substitute this back into the previous result: \(|x| = k |\frac{y}{x} - 1|\). Simplify to obtain the implicit solution \(x = k (\frac{y-x}{x})\) or \(x = \frac{y-x}{c}\), indicating that for some constant \(C\), \(x = \frac{y-x}{c}\).

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Key Concepts

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

Method of Substitution
In the context of solving differential equations, the method of substitution is a powerful technique to simplify and solve equations that at first glance appear complicated. By strategically introducing new variables, substitution converts a differential equation into a simpler form, often a recognizably solvable type.

Specifically for homogeneous differential equations, one common substitution is setting a term, such as the dependent variable divided by the independent variable, equal to a new variable. In our exercise example, the substitution technique involves replacing the variable 'y' with 'vx', where 'v' is a new variable function of 'x'. This substitution works because it exploits the nature of homogeneous equations, where all terms can be rewritten as a function of a single variable.

Substituting 'y = vx' simplifies the original differential equation into a more manageable form by aligning terms with respect to this new variable. This approach leads to a transformed equation that is easier to solve using methods such as variable separation. It's an effective strategy as it leverages the properties of homogeneous functions to reduce complexity.
Differential Equation Solutions
Solving differential equations is an essential part of mathematics that helps us understand systems evolving over time or space. In our example, the solution process begins by transforming the given equation into a format that is more accessible for standard solving techniques.

By applying substitution, the original equation becomes a simpler homogeneous equation: \((1-v) \, dx + x \, dv = 0\). This newly formed equation can be addressed using standard integration techniques.
  • First, differentiate the substituted function to fit the new equation format.
  • Simplify and rearrange the equation to isolate terms properly.
  • Finally, integrate both sides to obtain the general solution.

The resulting solution is an expression, or "family" of solutions, typically involving an arbitrary constant, which encompasses all possible behaviors of the system under initial conditions. The process shows the power of substitution not only to simplify equations but to systematically reach a solution set.
Variable Separation
Variable separation is one of the simplest yet most effective ways to solve differential equations. It involves rearranging the terms of the differential equation so that each variable and its corresponding differential are on opposite sides of the equation.

In the context of the exercise, we achieve variable separation after simplifying and manipulating the equation into \(\frac{dx}{x} = \frac{dv}{v-1}\). This format allows us to perform separate integrals on each side, essentially integrating with respect to their own variables.

  • Left side \(\frac{dx}{x}\) is integrated to give \(\ln|x|\).
  • Right side \(\frac{dv}{v-1}\) leads to \(\ln|v-1|\).

This method turns a complex problem into two independent and simpler problems, easily solved by basic integration. The clear separation between variables simplifies the integration process, aiding in the derivation of a general solution that can be transformed back into terms of the original variables.

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Most popular questions from this chapter

Radioactive Decay Series The following system of differential equations is encountered in the study of the decay of a special type of radioactive series of elements: $$ \begin{aligned} &\frac{d x}{d t}=-\lambda_{1} x \\ &\frac{d y}{d t}=\lambda_{1} x-\lambda_{2} y \end{aligned} $$ where \(\lambda_{1}\) and \(\lambda_{2}\) are constants. Discusshow to solve this system subject to \(x(0)=x_{0}, y(0)=y_{0}\). Carry out your ideas.

A large tank is partially filled with 100 gallons of fluid in which 10 pounds of salt is dissolved. Brine containing \(\frac{1}{2}\) pound of salt per gallon is pumped into the tank at a rate of \(6 \mathrm{gal} / \mathrm{min}\). The well-mixed solution is then pumped out at a slower rate of \(4 \mathrm{gal} / \mathrm{min}\). Find the number of pounds of salt in the tank after 30 minutes.

Use a numerical solver and Euler'smethodto approximate \(y(1.0)\) where \(y(x)\) is the solution to \(y^{\prime}=2 x y^{2}, y(0)=1\). First use \(h=0.1\) and then \(h=0.05\). Repeat using the \(\mathrm{RK} 4\) method. Discuss what might cause the approximations of \(y(1.0)\) to differ so greatly.

Rocket Motion Suppose a small single-stage rocket of total mass \(m(t)\) is launched vertically and that the rocket consumes its fuel at a constant rate. If the positive direction is upward and if we take air resistance to be linear, then a differential equation for its velocity \(v(t)\) is given by $$ \frac{d v}{d t}+\frac{k-\lambda}{m_{0}-\lambda t} v=-g+\frac{R}{m_{0}-\lambda t^{\prime}} $$ where \(k\) is the drag coefficient, \(\lambda\) is the rate at which fuel is consumed, \(R\) is the thrust of the rocket, \(m_{0}\) is the total mass of the rocket at \(t=0\), and \(g\) is the acceleration due to gravity. See Problem 21 in Exercises \(1.3\). (a) Find the velocity \(v(t)\) of the rocket if \(m_{0}=200 \mathrm{~kg}\), \(R=2000 \mathrm{~N}, \lambda=1 \mathrm{~kg} / \mathrm{s}, g=9.8 \mathrm{~m} / \mathrm{s}^{2}, k=3 \mathrm{~kg} / \mathrm{s}\), and \(v(0)=0\) (b) Use \(d s / d t=v\) and the result in part (a) to find the height \(s(t)\) of the rocket at time \(t\).

(a) A simple model for the shape of a tsunami is given by $$ \frac{d W}{d x}=W \sqrt{4-2 W}, $$ where \(W(x)>0\) is the height of the wave expressed as a function of its position relative to a point offshore. By inspection, find all constant solutions of the \(\mathrm{DE}\). (b) Solve the differential equation in part (a). A CAS may be useful for integration. (c) Use a graphing utility to obtain the graphs of all solutions that satisfy the initial condition \(W(0)=2\).

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