Chapter 5: Problem 18
L?sen Sie die folgenden Differentialgleichungen 1 . Ordnung (gemischte Aufgaben): a) \(y^{\prime}=x\left(y^{2}+1\right)\) b) \(y^{\prime}=y \cdot \sin x\) c) \(y^{\prime}=x y\) d) \(x y^{\prime}+y=2 \cdot \ln x\) e) \(y^{\prime}=5 x^{4}(y+1)\) f) \(y^{\prime}-5 y=2 \cdot \cos x-\sin (3 x)\)
Short Answer
Expert verified
For each problem: a) \( y = \tan(\frac{x^2}{2} + C) \), b) \( y = Ce^{-\cos x} \), c) \( y = Ce^{ \frac{x^2}{2}} \), d) \( y = x \ln x - \frac{x}{2} + \frac{C}{x} \), e) \( y = De^{x^5} - 1 \), f) Use an integrating factor to solve.
Step by step solution
01
Solving (a) \( y' = x(y^2 + 1) \)
First, separate the variables: \( \frac{dy}{y^2 + 1} = x \, dx \). Then integrate both sides: \( \int \frac{dy}{y^2 + 1} = \int x \, dx \). This gives \( \arctan(y) = \frac{x^2}{2} + C \), where \( C \) is the constant of integration.
02
Solving (b) \( y' = y \sin x \)
Separate the variables: \( \frac{dy}{y} = \sin x \, dx \). Integrate both sides: \( \int \frac{dy}{y} = \int \sin x \, dx \). This results in \( \ln |y| = -\cos x + C \). Therefore, \( y = e^{C - \cos x} \) or \( y = Ce^{-\cos x} \) where \( C = e^{C} \).
03
Solving (c) \( y' = xy \)
Separate the variables: \( \frac{dy}{y} = x \, dx \). Integrate both sides: \( \int \frac{dy}{y} = \int x \, dx \). This gives \( \ln |y| = \frac{x^2}{2} + C \). Therefore, \( y = e^{\frac{x^2}{2} + C} \) or \( y = Ce^{\frac{x^2}{2}} \) where \( C = e^{C} \).
04
Solving (d) \( x y' + y = 2 \ln x \)
Rewrite the equation: \( y' + \frac{y}{x} = 2 \frac{\ln x}{x} \). This is a linear differential equation. Use the integrating factor \( \mu(x) = e^{\int \frac{1}{x} \, dx} = x \). Multiply both sides by the integrating factor: \( x y' + y = 2 x \ln x \). The left side can be rewritten as \( (xy)' = 2 x \ln x \). Integrate both sides: \( xy = \int 2 x \ln x \, dx \), \( xy = x^2 \ln x - \frac{x^2}{2} + C \). Therefore, \( y = x \ln x - \frac{x}{2} + \frac{C}{x} \).
05
Solving (e) \( y' = 5 x^4 (y + 1) \)
Separate the variables: \( \frac{dy}{y + 1} = 5 x^4 \, dx \). Integrate both sides: \( \int \frac{dy}{y + 1} = \int 5 x^4 \, dx \). This results in \( \ln |y + 1| = x^5 + C \). Therefore, \( y + 1 = e^{x^5 + C} \) or \( y = De^{x^5} - 1 \) where \( D = e^{C} \).
06
Solving (f) \( y' - 5y = 2 \cos x - \sin(3x) \)
This is a linear nonhomogeneous differential equation. The integrating factor is \( \mu(x) = e^{-5x} \). Multiply both sides by the integrating factor: \( e^{-5x} y' - 5e^{-5x} y = e^{-5x}(2 \cos x - \sin(3x)) \). The left side can be rewritten as \( (e^{-5x} y)' = e^{-5x}(2 \cos x - \sin(3x)) \). Integrate both sides: \( e^{-5x} y = \int e^{-5x}(2 \cos x - \sin(3x)) \, dx \). Solutions involve integration by parts; solve the integrals to find \( y \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Order Differential Equations
First order differential equations are equations that involve the first derivative of a function but no higher derivatives. They can be represented in the form:
Some common types of first order differential equations include:
- \( \frac{dy}{dx} = f(x, y) \)
Some common types of first order differential equations include:
- Separable equations
- Linear equations
- Exact equations
- Homogeneous equations
- Bernoulli equations
Separation of Variables
The technique of separation of variables is a powerful method for solving first order differential equations. It involves rearranging the equation so that each variable appears on a different side of the equation. The goal is to express the differential equation in the form:
Here is a step-by-step approach for solving using separation of variables:
- \( \frac{dy}{dx} = g(x)h(y) \)
- \( \frac{1}{h(y)}dy = g(x)dx \)
Here is a step-by-step approach for solving using separation of variables:
- Rewrite the equation: Move all terms involving one variable (y) to one side, and all terms involving the other variable (x) to the other side.
- Integrate both sides: Integrate the equation with respect to their respective variables.
- Add constant of integration: After integrating, add the constant of integration (C). This often appears after the integration process and varies depending on the equation.
- Solve for y: If possible, solve the resulting expression for the dependent variable y.
Linear Differential Equations
Linear differential equations are a specific type of differential equation where the dependent variable and its derivatives appear linearly. The general form of a first-order linear differential equation is:
- \( y' + P(x)y = Q(x) \)
- Identify P(x): Note the coefficient function of y, denoted as \( P(x) \).
- Compute the integrating factor \( \, \mu(x) \): This factor is given by \( \mu(x) = e^{ \, \int P(x)dx} \).
- Multiply the differential equation by \( \, \mu(x) \): This will transform the left side of the equation into the derivative of a product.
- Integrate both sides: The equation now transforms into \( d/dx[ \mu(x)y] = \mu(x)Q(x) \). Integrate both sides with respect to x.
- Solve for y: Solve the resulting equation for y to find the general solution.
Integrating Factor
An integrating factor is a function used to simplify the process of solving linear first-order differential equations. The integrating factor, usually denoted as \( \,\mu(x) \), helps transform the differential equation into an exact equation, making it easier to integrate. To find the integrating factor, follow these steps:
- Write the equation in standard form: Ensure the differential equation is in the form \( y' + P(x)y = Q(x) \).
- Calculate \( \,\mu(x) \): Compute the integrating factor using the formula \( \,\mu(x) = e^{ \int P(x)dx} \).
- Multiply through by \( \,\mu(x) \): Apply the integrating factor to both sides of the equation to obtain \( \mu(x)y' + \mu(x)P(x)y = \mu(x)Q(x) \).
- Simplify the left-hand side: Recognize that the left-hand side of the equation is now \( \frac{d}{dx}[ \mu(x)y] \, \), which can be integrated directly.
- Integrate both sides: Integrate the simplified equation with respect to x to find an expression for y.