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Chapter 2: Problem 27

Solve the given differential equation by using an appropriate substitution. $$\frac{d y}{d x}=2+\sqrt{y-2 x+3}$$

Short Answer

Expert verified
Substitute \( v = y - 2x + 3 \), solve \( v \), and integrate to find \( y = 2x - 3 + \left( \frac{x + C}{2} \right)^2 \).

Step by step solution

01

Identify a Suitable Substitution

To solve the given differential equation, notice the term \( \sqrt{y - 2x + 3} \). It suggests a substitution could simplify the expression. Let \( v = y - 2x + 3 \). Then, \( y = v + 2x - 3 \) and the differential \( \frac{dy}{dx} = \frac{dv}{dx} + 2 \).
02

Substitute into the Differential Equation

Replace \( y \) and \( \frac{dy}{dx} \) in the original differential equation. We get:\[ \frac{dv}{dx} + 2 = 2 + \sqrt{v} \]Now simplify the equation:\[ \frac{dv}{dx} = \sqrt{v} \]
03

Separate Variables

The equation \( \frac{dv}{dx} = \sqrt{v} \) allows us to separate variables. Rearrange to:\[ \frac{dv}{\sqrt{v}} = dx \]
04

Integrate Both Sides

Integrate both sides of the separated equation:\[ \int \frac{dv}{\sqrt{v}} = \int dx \]The left-hand side integrates as \( 2\sqrt{v} \) and the right-hand side as \( x + C \):\[ 2\sqrt{v} = x + C \]
05

Solve for the Original Variables

From \( 2\sqrt{v} = x + C \), solve for \( v \):\[ v = \left( \frac{x + C}{2} \right)^2 \]Recall the substitution \( v = y - 2x + 3 \):\[ y - 2x + 3 = \left( \frac{x + C}{2} \right)^2 \]Finally, solve for \( y \):\[ y = 2x - 3 + \left( \frac{x + C}{2} \right)^2 \]
06

Final Answer

The solution to the differential equation \( \frac{d y}{d x}=2+\sqrt{y-2 x+3} \) is:\[ y = 2x - 3 + \left( \frac{x + C}{2} \right)^2 \]

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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 tool used in solving differential equations, especially when they contain complex expressions. In this approach, we try to simplify the equation by introducing a new variable.
For instance, when faced with the differential equation \( \frac{d y}{d x}=2+\sqrt{y-2 x+3} \), the expression \( \sqrt{y - 2x + 3} \) suggests a substitution.
  • We set \( v = y - 2x + 3 \), helping to transform the equation into a more manageable form.
  • This choice effectively turns the original equation into a simpler one: \( \frac{dv}{dx} = \sqrt{v} \).
This substitution reduces complexity and enables easier manipulation of the equation, paving the way for further problem-solving methods such as separation of variables.
Variable Separation
In solving differential equations, variable separation is an essential technique. This means rearranging the equation so each variable appears on opposite sides of the equation.
For the equation \( \frac{dv}{dx} = \sqrt{v} \), separating variables involves isolating \( v \) and \( x \).
  • We write: \( \frac{dv}{\sqrt{v}} = dx \).
  • Here, \( v \) terms are on one side and \( x \) terms on the other.
This separation is crucial, as it prepares the equation for integration. By isolating the variables, we align them ready for the next step toward finding the solution.
Integration
Integration is the process used to reverse differentiation, allowing us to solve differential equations after separation of variables.
After separating the variables, we integrate both sides of the equation: \( \int \frac{dv}{\sqrt{v}} = \int dx \).
  • The left side integrates to \( 2\sqrt{v} \), converting the differentiable component back to its integral form.
  • The right side integrates to \( x + C \), where \( C \) is the constant of integration.
Integration confirms our equation's transformation and allows us to move toward solving for the original variables. It's a crucial step for culminating the simplification process begun by substitution.
Differential Equation Solution Steps
Solving a differential equation often involves several methodical steps. Understanding each step provides clarity:
  • **Identify a Suitable Substitution:** Choose a substitution that simplifies the equation, like \( v = y - 2x + 3 \), which transforms the equation's complexity.
  • **Substitute into the Differential Equation:** Insert the substitution back into the equation, resulting in \( \frac{dv}{dx} = \sqrt{v} \).
  • **Separate Variables:** Rearrange the equation so each variable is isolated, as seen in \( \frac{dv}{\sqrt{v}} = dx \).
  • **Integrate Both Sides:** Perform integration on both sides to solve for \( v \) in terms of \( x \), resulting in \( 2\sqrt{v} = x + C \).
  • **Solve for the Original Variables:** Replace \( v \) with its original terms, revert to the original variable \( y \), yielding the solution \( y = 2x - 3 + \left( \frac{x + C}{2} \right)^2 \).
By following these guided steps, tackling complex differential equations becomes systematic and understandable.

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Most popular questions from this chapter

Solve the given initial-value problem. $$\frac{d y}{d x}=\cos (x+y), \quad y(0)=\pi / 4$$

Often a radical change in the form of the solution of a differential equation corresponds to a very small change in either the initial condition or the equation itself. Find an explicit solution of the given initial-value problem. Use a graphing utility to plot the graph of each solution. Compare each solution curve in a neighborhood of (0,1). $$\frac{d y}{d x}=(y-1)^{2}, \quad y(0)=1$$

Differential equations are sometimes solved by having a clever idea. Here is a little exercise in cleverness: Although the differential equation \((x-\sqrt{x^{2}+y^{2}}) d x+y d y=0\) is not exact, show how the rearrangement \((x d x+y d y) / \sqrt{x^{2}+y^{2}}=d x\) and the observation \(\frac{1}{2} d\left(x^{2}+y^{2}\right)=x d x+y d y\) can lead to a solution.

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. Give the largest interval \(I\) over which the solution is defined. $$x \frac{d y}{d x}+y=4 x+1, \quad y(1)=8$$
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