Chapter 7: Problem 11
Solve each differential equation. $$ \frac{d y}{d x}-\frac{y}{x}=3 x^{3} ; y=3 \text { when } x=1 $$
Short Answer
Expert verified
The solution to the differential equation is \( y = x^4 + 2x \).
Step by step solution
01
Identify the type of differential equation
The given equation is \( \frac{dy}{dx} - \frac{y}{x} = 3x^3 \), which is a first-order linear differential equation. We can express it in the standard form: \( \frac{dy}{dx} + P(x)y = Q(x) \) where \( P(x) = -\frac{1}{x} \) and \( Q(x) = 3x^3 \).
02
Find the integrating factor
The integrating factor, \( \mu(x) \), is found using \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = -\frac{1}{x} \), so \( \mu(x) = e^{-\int \frac{1}{x} \, dx} = e^{-\ln|x|} = \frac{1}{x} \).
03
Multiply through by the integrating factor
Multiply the entire differential equation by the integrating factor \( \frac{1}{x} \) to get: \( \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = 3x^2 \).
04
Simplify to a derivative of a product
Recognize the left side as a derivative of a product: \( \frac{d}{dx}(\frac{y}{x}) = 3x^2 \).
05
Integrate both sides
Integrate both sides of the equation with respect to \( x \): \( \int \frac{d}{dx}(\frac{y}{x}) \, dx = \int 3x^2 \, dx \). This gives \( \frac{y}{x} = x^3 + C \), where \( C \) is the constant of integration.
06
Solve for \( y \)
Multiply through by \( x \) to solve for \( y \): \( y = x^4 + Cx \).
07
Use the initial condition to find \( C \)
Substitute the initial condition \( y = 3 \) when \( x = 1 \) into \( y = x^4 + Cx \) to find \( C \). Plugging the values in gives: \( 3 = 1^4 + C(1) \). Thus, \( C = 2 \).
08
Write the particular solution
Substitute \( C = 2 \) into the equation \( y = x^4 + Cx \) to get the specific solution: \( y = x^4 + 2x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equation
In mathematics, understanding differential equations is crucial for modeling many real-world phenomena. A first-order linear differential equation is one of the fundamental types of differential equations you'll encounter. It generally takes the form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
Here, the equation features a derivative of the unknown function \( y \) with respect to \( x \), along with terms that possibly involve \( y \) and \( x \) themselves. The coefficients \( P(x) \) and \( Q(x) \) are functions of \( x \).
Key points about first-order linear differential equations:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
Here, the equation features a derivative of the unknown function \( y \) with respect to \( x \), along with terms that possibly involve \( y \) and \( x \) themselves. The coefficients \( P(x) \) and \( Q(x) \) are functions of \( x \).
Key points about first-order linear differential equations:
- They involve the first derivative \( \frac{dy}{dx} \) and no higher derivatives.
- The function \( y \) is linear, meaning it appears without exponents other than 1.
- These equations can often be solved using special techniques, such as integrating factors.
Integrating Factor
An integrating factor is a useful mathematical tool for solving first-order linear differential equations. It simplifies the differential equation, making it easier to solve.
The integrating factor, usually denoted as \( \mu(x) \), is a function you multiply throughout the equation turning the left side into an exact derivative:
\[ \mu(x) = e^{\int P(x) \, dx} \]
In this equation, \( P(x) \) is the coefficient of \( y \) from the standard form of the differential equation. This transformation enables us to rearrange the equation into something that can be integrated more straightforwardly.Steps to use an integrating factor:
The integrating factor, usually denoted as \( \mu(x) \), is a function you multiply throughout the equation turning the left side into an exact derivative:
\[ \mu(x) = e^{\int P(x) \, dx} \]
In this equation, \( P(x) \) is the coefficient of \( y \) from the standard form of the differential equation. This transformation enables us to rearrange the equation into something that can be integrated more straightforwardly.Steps to use an integrating factor:
- Identify \( P(x) \).
- Calculate the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).
- Multiply the entire differential equation by \( \mu(x) \).
- The equation now becomes easily integrable on the left side with respect to \( x \).
Initial Condition
In solving differential equations, an initial condition provides the specific values that allow us to find a unique solution. This is necessary because differential equations typically have infinitely many solutions.
An initial condition will usually specify the value of the unknown function \( y \) at a certain point \( x_0 \), such as \( y = 3 \) when \( x = 1 \).How initial conditions work:
An initial condition will usually specify the value of the unknown function \( y \) at a certain point \( x_0 \), such as \( y = 3 \) when \( x = 1 \).How initial conditions work:
- They help in calculating the constant of integration \( C \) after solving the differential equation.
- Using the initial condition, you substitute these values into your general solution.
- Solve the equation for \( C \) to get the particular solution that fits the given condition.