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Chapter 2: Problem 9

Solve the given differential equation by using an appropriate substitution. $$-y d x+(x+\sqrt{x y}) d y=0$$

Short Answer

Expert verified
Use substitution \( y = vx \), simplify, integrate, and back substitute to find the relationship between \( x \) and \( y \).

Step by step solution

01

Identify the Type of Differential Equation

Our differential equation is \[-y \ dx +(x+\sqrt{x y}) \ dy=0\]This resembles a Homogeneous Differential Equation because it involves terms that can potentially simplify to a form that depends only on \( \frac{y}{x} \). We will employ a substitution to leverage this property.
02

Use Appropriate Substitution

The standard substitution for a homogeneous differential equation is \( y = vx \), where \( v \) is a function of \( x \). Differentiate this to get \( dy = v \ dx + x \ dv \). Substitute \( y = vx \) and \( dy \) into the original equation:\[-v x \ dx + (x + \sqrt{x(vx)}) (v \ dx + x \ dv) = 0\]
03

Simplify the Equation

Simplifying inside the square root, we get \( \sqrt{x^2 v} = x \sqrt{v} \). Inserting this back into the equation results in:\[-v x \ dx + \left( x + x\sqrt{v} \right) (v \ dx + x \ dv) = 0\]Expanding the terms inside the equation gives:\[-v x \ dx + (x v + x \sqrt{v}) \ dx + (v x^2 + x^2 \sqrt{v}) \ dv = 0\]
04

Combine Like Terms

Grouping like terms gives:\[(x v + x \sqrt{v} - v x) \ dx + (v x^2 + x^2 \sqrt{v}) \ dv = 0\]The \( dx \) terms simplify to \( x \sqrt{v} \ dx \), resulting in:\[x \sqrt{v} \ dx + x^2 (v + \sqrt{v}) \ dv = 0\]
05

Divide by Common Factor

Factor out a common factor from the equation:\[\sqrt{v} \ dx + x (v + \sqrt{v}) \ dv = 0\]Divide the whole equation by \( x \sqrt{v} \):\[\frac{1}{x} \ dx + \frac{v + \sqrt{v}}{\sqrt{v}} \ dv = 0\]
06

Integrate Both Sides

This equation separates naturally:\[\frac{1}{x} \ dx = - \frac{v + \sqrt{v}}{\sqrt{v}} \ dv = - (\sqrt{v} + 1) \ dv\]Integrate both sides:\[\int \frac{1}{x} \, dx = \int - (\sqrt{v} + 1) \, dv\]This yields:\[\ln|x| = -\frac{2}{3}v^{3/2} - v + C\]
07

Back Substitute

Since \( v = \frac{y}{x} \), substitute back \( v = \frac{y}{x} \) into the equation:\[\ln|x| = -\frac{2}{3}\left(\frac{y}{x}\right)^{3/2} - \frac{y}{x} + C\]
08

Simplify to the Final Solution

Move terms and simplify if necessary. The final form of the solution incorporates the relationship between \( x \) and \( y \) described by the differential equation specific to the constants of integration.

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Key Concepts

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

Differential Equations
Differential equations involve mathematical expressions that relate a function with its derivatives. These are crucial in various fields as they describe how things change over time or space. In this particular case, we're dealing with a first-order differential equation:
  • First-order because it involves only the first derivative, not higher ones.
  • Date back to modeling processes like heat conduction, wave motion, or growth decay.
Homogeneous differential equations are a specific type. They often allow for simpler solutions because they maintain the same degree of homogeneity across terms. This means if you shock the system by changing all variables by a multiplier, the system changes in a consistent way. Understanding this property can help us find solutions with new variables.
Variable Substitution
Variable substitution is a handy tool when solving complex differential equations. It involves replacing a variable with another expression, making it easier to work with.
  • Here, we use the substitution \( y = vx \), where \( v \) is a new variable.
  • This approach leverages the homogeneity to simplify terms.
  • By introducing \( v \), our problem transforms into one involving \( v \) and \( x \), turning our original equation into a more manageable form.
  • This transformation makes differentiating and integrating smoother as variables become more independent.
This technique highlights the power of adapting equations to reveal simpler structures.
Integration Techniques
Integration is a fundamental tool to reverse the process of differentiation. Once we've transformed our differential equation using substitution, integration helps solve the equation. In this exercise:
  • Our equation became separable, meaning we could rewrite it with each variable isolated on different sides.
  • We used the integral \( \int \frac{1}{x} \, dx = \ln|x| \) for the \( x \) terms.
  • The \( v \) terms included \( \int - (\sqrt{v} + 1) \, dv \), integrating both requires understanding powers and roots.
Integration is the step where we systematically unravel a problem by summing infinitely small areas to find accumulated quantities. This allows us to make sense of change across variables.
Mathematical Modeling
Mathematical modeling refers to representing real-world phenomena using mathematical expressions. The differential equation in the exercise could represent a dynamic system in engineering, physics, or economics. By expressing such systems with equations:
  • We can forecast outcomes or describe patterns over time.
  • Differential equations often capture the rate of change, offering insight into the evolution of a system.
This specific problem involved designing a solution through substitutions and integrations, reflecting the type of precise thinking needed in mathematical modeling. Such modeling inspires problem-solvers to think creatively and strategically, steering through complexities.

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Most popular questions from this chapter

Heart Pacemaker A heart pacemaker consists of a switch, a battery of constant voltage \(E_{0},\) a capacitor with constant capacitance \(C,\) and the heart as a resistor with constant resistance \(R\). When the switch is closed, the capacitor charges; when the switch is open, the capacitor discharges, sending an electrical stimulus to the heart. During the time the heart is being stimulated, the voltage \(E\) across the heart satisfies the linear differential equation $$\frac{d E}{d t}=-\frac{1}{R C} E$$ Solve the DE, subject to \(E(4)=E_{0}\).

Solve the given initial-value problem. $$(x+y)^{2} d x+\left(2 x y+x^{2}-1\right) d y=0, \quad y(1)=1$$

Solve the given differential equation by using an appropriate substitution. $$\frac{d y}{d x}=1+e^{y-x+5}$$

Determine whether the given differential equation is exact. If it is exact, solve it. $$\left(x^{2} y^{3}-\frac{1}{1+9 x^{2}}\right) \frac{d x}{d y}+x^{3} y^{2}=0$$

Solve the given initial-value problem. $$x^{2} \frac{d y}{d x}-2 x y=3 y^{4}, \quad y(1)=\frac{1}{2}$$
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