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Chapter 31: Problem 28

Solve the given differential equations. \(y^{\prime}+y=y^{2}\) (Solve by letting \(y=1 / u\) and solving the resulting linear equation for \(u\).)

Short Answer

Expert verified
The solution is \( y = \frac{1}{1 + Ce^{t}} \).

Step by step solution

01

Substitute y

Given the differential equation is \( y' + y = y^2 \). Substitute \( y = \frac{1}{u} \) to transform the equation. This implies \( y' = -\frac{1}{u^2} \cdot u' \).
02

Apply the substitution

Substituting \( y = \frac{1}{u} \) and \( y' = -\frac{1}{u^2} \cdot u' \) into the original equation gives us: \( -\frac{1}{u^2} \cdot u' + \frac{1}{u} = \frac{1}{u^2} \).
03

Simplify the equation

Multiply the entire equation by \( u^2 \) to clear the fractions: \(-u' + u = 1\).
04

Rearrange into standard form

Rearrange the equation \(-u' + u = 1\) to the standard linear differential equation form \( u' - u = -1 \).
05

Find the integrating factor

The standard form is \( u' + Pu = Q \), where \( P = -1 \). Find the integrating factor: \( \mu(t) = e^{\int -1 dt} = e^{-t} \).
06

Apply the integrating factor

Multiply through by the integrating factor \( e^{-t} \): \( e^{-t}u' - e^{-t}u = -e^{-t} \).
07

Simplify using the integrating factor

Notice that the left side is \( \frac{d}{dt}(e^{-t}u) = -e^{-t} \). Integrate both sides: \( e^{-t}u = \int -e^{-t} dt \).
08

Integrate and solve for u

Integrate the right side: \( e^{-t}u = -\int e^{-t} dt = -(-e^{-t}) = e^{-t} + C \) (where \( C \) is the constant of integration).
09

Solve for u

Solving for \( u \), divide through by \( e^{-t} \): \( u = 1 + Ce^{t} \).
10

Substitute back for y

Since \( y = \frac{1}{u} \), substitute back to obtain: \( y = \frac{1}{1 + Ce^{t}} \). This is the solution to the original differential equation.

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Key Concepts

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

Integrating Factor
The integrating factor is a key technique often used to solve first-order linear differential equations. Understanding it can greatly simplify the process of finding solutions.

An integrating factor is a function, typically represented by \( \mu(t) \), which when multiplied to the given differential equation makes the left side easily transformable into a derivative.
  • The general form of a first-order linear differential equation is \( y' + P(t)y = Q(t) \).
  • The integrating factor \( \mu(t) \) is defined as \( e^{\int P(t) dt} \).
  • Multiplying through the entire equation by the integrating factor helps us express the left-hand side as a derivative, leading to \( \frac{d}{dt}(\mu(t)y) = \mu(t)Q(t) \).
This method eventually allows us to integrate both sides, simplifying to solve for the function \( y(t) \).
Substitution Method
Substitution is a technique used to simplify differential equations by transforming their form, enabling easier solutions.

In the given exercise, substituting \( y = \frac{1}{u} \) helped change a non-linear equation to a linear one.
  • This new substitution leads us to differentiate \( y \) in terms of \( u \), finding \( y' \) and its relation to \( u' \).
  • Substituting these expressions into the original equation results in a simpler form, which can often be rewritten as a standard linear equation.
By using the substitution method, we simplify the solving process, often converting a more complex differential equation into a manageable one.
Linear Differential Equation
Linear differential equations are a class of equations involving unknown functions and their derivatives.

The standard form of a first-order linear differential equation is \( y' + P(t)y = Q(t) \). This indicates:
  • \( P(t) \) and \( Q(t) \) are continuous functions that depend on the variable \( t \).
  • The equation is linear concerning both \( y \) and \( y' \), ensuring no products or powers of \( y \) and \( y' \) appear.
Linear differential equations can be systematically solved, often by applying methods such as the integrating factor, which facilitates integration and solution building.
Constant of Integration
When integrating to solve differential equations, a constant of integration \( C \) is included to represent the family of solutions.

The constant of integration arises because antiderivatives are not unique, expressing the idea that there are infinitely many functions whose derivative is a given function.
  • For instance, integrating \( f'(x) \) to find \( f(x) \) yields an arbitrary constant \( C \), illustrating any shift upwards or downwards on a graph.
  • In differential equations, \( C \) helps encapsulate all possible solutions that satisfy the equation.
When solving or working with initial value problems, \( C \) can be determined specifically using given conditions or additional information, personalizing the solution to the scenario.

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Most popular questions from this chapter

Solve the given problems by solving the appropriate differential equation. In a ballistics test, a bullet is fired into a sandbag. The acceleration of the bullet within the sandbag is \(-40 \sqrt{v},\) where \(v\) is the velocity (in \(\mathrm{ft} / \mathrm{s}\) ). When will the bullet stop if it enters the sandbag at \(950 \mathrm{ft} / \mathrm{s} ?\)

Solve the given differential equations by Laplace transforms. The function is subject to the given conditions. A 20 -mH inductor, a \(40-\Omega\) resistor, a \(50-\mu \mathrm{F}\) capacitor, and a voltage source of \(100 e^{-1000 t}\) are connected in series in an electric circuit. Find the charge on the capacitor as a function of time \(t\), if \(q=0\) and \(i=0\) when \(t=0\)

Find the inverse transforms of the given functions of \(s\). $$F(s)=\frac{1}{s^{3}+3 s^{2}+3 s+1}$$

Solve the given differential equations by Laplace transforms. The function is subject to the given conditions. $$2 y^{\prime \prime}+8 y=3 \sin 2 t, y(0)=0, y^{\prime}(0)=0$$

Solve the given differential equations by Laplace transforms. The function is subject to the given conditions. $$2 y^{\prime \prime}+y^{\prime}-y=\sin 3 t, \quad y(0)=0, y^{\prime}(0)=0$$
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