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Chapter 2: Problem 30

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

Short Answer

Expert verified
Use substitution \( z = 3x + 2y \), integrate, apply initial conditions, and solve for the constant.

Step by step solution

01

Analyze and Simplify the Equation

Observe the differential equation \( \frac{d y}{d x}=\frac{3 x+2 y}{3 x+2 y+2} \). Notice that it is a first-order differential equation with variables \(x\) and \(y\). The right-hand side seems complex, but it's a form where separation or substitution might simplify the problem.
02

Substitute to Simplify

Let \( z = 3x + 2y \), which gives us \( \, \frac{d z}{dx} = 3 + 2 \frac{dy}{dx} \, \). The original equation becomes \( \, \frac{d y}{d x} = \frac{z}{z+2} \,\). Therefore, \( \, \frac{d z}{dx} = 3 + 2(\frac{z}{z+2}) \,\).
03

Simplify Substituted Equation

Simplify \( \, \frac{d z}{dx} = 3 + 2 \frac{z}{z+2} \,\) to get \( \, \frac{d z}{dx} = 3 + \frac{2z}{z+2} \,\). Simplify further: \( \, \frac{d z}{dx} = \frac{3(z+2) + 2z}{z+2} = \frac{5z + 6}{z+2} \,\).
04

Separate Variables

Rearrange terms to separate variables: \( \, \frac{(z+2)}{5z + 6} \frac{dz}{dx} = 1 \,\). This allows integration: \( \, rac{(z+2)}{5z + 6} \, dz = dx \,\).
05

Integrate Both Sides

Integrate the left side with respect to \( z \) and the right side with respect to \( x \): \[ \int \frac{z+2}{5z + 6} \, dz = \int 1 \, dx. \] Solve the integral: transform the left side using partial fraction decomposition or another method if direct integration is complex.
06

Apply Initial Condition

After finding the integrated function (let's call it \( F(z) = x + C \)), apply the initial condition \( y(-1) = -1 \) to find \( C \). Substitute \( x = -1 \) and \( y = -1 \) into the solution equation to solve for the constant. Use \( z = 3x + 2y = -3 + -2 = -5 \).
07

Solve for C and Conclude

Substitute back to solve for \( C \) in \( F(-5) = -1 + C \). Once \( C \) is known, substitute back \( z = 3x + 2y \) to get a relation between \( y \) and \( x \). This gives the explicit solution to the differential equation.

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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 clever technique to simplify complex differential equations, especially first-order ones. When faced with an equation like \( \frac{dy}{dx} = \frac{3x+2y}{3x+2y+2} \), directly solving it might be challenging. To simplify this, we can use a substitution.
  • Choose a substitution that conveniently replaces part of the equation with a new variable, often denoted as \( z \).
  • In our example, we substitute \( z = 3x + 2y \). This substitution helps to condense the expression, reducing the complexity.
  • After substitution, we derive new expressions in terms of \( z \) and \( x \), facilitating the separation of variables or straightforward integration.
This method relies heavily on recognizing parts of the equation that can be grouped and substituted, leading to easier manipulation and solving.
Separation of Variables
Separation of variables is a technique that allows us to solve differential equations by rewriting them so that each variable can be integrated independently. For example, after substitution, our problem becomes \( \frac{dz/dx}{5z + 6} = \frac{1}{z+2} \).
  • This manipulation enables us to rearrange terms so one side only contains \( z \) terms and the other \( x \) terms.
  • In the transformed expression \( \frac{(z+2)}{5z + 6} \frac{dz}{dx} = 1 \), integration becomes possible once both sides are expressed in terms of differentials involving only one variable each.
  • This method significantly simplifies the integration process, allowing us to derive the integrated forms on both sides easily.
By separating variables, we can effectively address initial value problems too, as the integration provides an explicit form to which initial conditions can be applied.
Initial Value Problems
Initial value problems require solving a differential equation under specified conditions at a particular point. In the example \( y(-1) = -1 \), this condition allows us to determine any constants arising from integration during the solution process.
  • Once the integration of the separated equation is performed, it generally includes a constant \( C \).
  • Applying the initial condition (here, substituting \( x = -1 \) and \( y = -1 \)), we can solve for \( C \).
  • In this exercise, the substitution simplifies \( z = 3x + 2y \) at \( x = -1 \) and \( y = -1 \), resulting in \( z = -5 \).
Knowing \( C \) is crucial as it particularizes the solution to fit the initial conditions, ensuring the solution is complete and specific to the given scenario. Integrating both sides in terms of separated variables and utilizing the initial condition will always yield a solution tailored to the problem's requirements.

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Most popular questions from this chapter

The population of a town grows at a rate proportional to the population present at time \(t\). The initial population of 500 increases by \(15 \%\) in 10 years. What will the population be in 30 years? How fast is the population growing at \(t=30 ?\)

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ x y^{\prime}+y=e^{x}, \quad y(1)=2 $$

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Devise an appropriate substitution to solve $$ x y^{\prime}=y \ln (x y) $$

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