Chapter 8: Problem 19
Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$\sin ^{2} x d y+\left[\sin ^{2} x+(x+y) \sin 2 x\right] d x=0$$
Short Answer
Expert verified
x + y = C|cosec^{2}x|
Step by step solution
01
- Rewrite the equation in standard form
The given differential equation is \[ \sin ^{2} x d y + \left[ \sin ^{2} x+(x+y) \sin 2 x \right] d x = 0 \]. Rewrite this as \[ \sin ^{2} x d y = - \left[ \sin ^{2} x+(x+y) \sin 2 x \right] d x \]. Dividing both sides by \sin ^{2} x, we get \[ d y = - \left[ 1 + (x + y) \frac{\sin 2 x}{\sin^2 x} \right] d x \]. Simplify the right-hand side: using \sin 2x = 2\sin x \cos x, it becomes \[ d y = - \left[ 1 + 2(x + y) \frac{\cos x}{\sin x} \right] d x \], which can be further simplified to \[ d y = - \left[ 1 + 2(x + y) \cot x \right] d x \].
02
- Identify the type of differential equation
Now, the differential equation is in the form \[ d y = - \left[ 1 + 2(x + y) \cot x \right] d x \]. This is a first-order, non-linear differential equation. However, observe that it can be put in the form of a linear first-order differential equation if shifted properly. Let's rewrite it with separation of variables in mind.
03
- Make the substitution
Let: \[ u = x + y \]. Then, \[ du = dx + dy \]. So, \[ dy = du - dx \]. Substituting back into the equation, we obtain\[ du - dx = -\left[ 1 + 2u \cot x \right] d x \]. Rearranging terms, \[ du = - dx - 2u \cot x \ dx \].
04
- Separate the variables
The equation can now be written as: \[ \left( du + 2u \cot x \ dx \right) = - dx \]. This simplifies to: \( \frac{du}{u} = -2 \cot x \ dx - \frac{dx}{u} \times constant term \), which isolates to: \( du = -asinx dx \) This can be integrated on both sides!
05
- Integrate both sides
Integrating both sides with respect to their variables, we have: \[ \int \frac{1}{u} du = -2 \int \cot x dx \]. Use constants: u , x and simplify
06
- Simplify the integrals
The integral of \frac{1}{u} du is \[ \ln |u| \], and the integral of \cot x dx is \[ \ln |\sin x| \], multiplied by -2 gives \[ -2 \ln |\sin x| \]. Therefore, we have \[ \ln | u | = -2 \ln | \sin x | + C \], with \C being the constant of integration. Rewriting, we get \[ \ln | x + y | = + \ln |\sin x | + \sin \]. Rearranged!
07
- Find the general solution
Rewrite the resulting equation in terms of u:\[ u = \frac{C}{\sin^{2}x} \] Substituting back \u = x + y: \[ x + y = C |\cosec^{2} x \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
linear first-order differential equation
Linear first-order differential equations have the general form \( \frac{dy}{dx} + P(x)y = Q(x) \). In these equations, the first-order derivative \( \frac{dy}{dx} \) is linear, which means \( y \) and its derivatives are raised only to the first power. The coefficients \( P(x) \) and \( Q(x) \) are functions of \( x \) only.
For example, consider a differential equation that can be written as \( \frac{dy}{dx} + P(x)y = Q(x) \). To solve this, one can use an integrating factor \( \rho(x) = e^{\bar{P(x) dx}} \), which converts the original equation into a solvable form. After multiplying the entire equation by the integrating factor, it can be integrated directly on both sides.
This concept is fundamental because it allows many real-world problems to be modeled and solved using linear equations.
For example, consider a differential equation that can be written as \( \frac{dy}{dx} + P(x)y = Q(x) \). To solve this, one can use an integrating factor \( \rho(x) = e^{\bar{P(x) dx}} \), which converts the original equation into a solvable form. After multiplying the entire equation by the integrating factor, it can be integrated directly on both sides.
This concept is fundamental because it allows many real-world problems to be modeled and solved using linear equations.
non-linear differential equations
Non-linear differential equations are equations where the unknown function and its derivatives appear non-linearly. This means that the function and its derivatives can be squared, cubed, or raised to some other power. They are generally more difficult to solve than linear equations. For instance, an equation of the form \( \frac{dy}{dx} = y^2 + x \) is non-linear.
Non-linear differential equations are prevalent in modeling complex systems like weather patterns, population dynamics, and fluid dynamics. These types of equations often require special methods of solution, which can include approximate, numerical, or graphical methods. Transformations and substitutions can sometimes simplify them into linear form or use specific techniques like separation of variables if they adhere to certain properties.
Non-linear differential equations are prevalent in modeling complex systems like weather patterns, population dynamics, and fluid dynamics. These types of equations often require special methods of solution, which can include approximate, numerical, or graphical methods. Transformations and substitutions can sometimes simplify them into linear form or use specific techniques like separation of variables if they adhere to certain properties.
separation of variables
Separation of variables is a method to solve differential equations. It involves rearranging the equation so that each variable and its differential is on opposite sides of the equation. For the differential equation \( \frac{dy}{dx} = g(x)h(y) \), it can be rewritten as \( \frac{1}{h(y)} dy = g(x) dx \).
Once separated, each side can be integrated independently:
Once separated, each side can be integrated independently:
- Integrate \( \frac{1}{h(y)} dy \) with respect to \( y \)
- Integrate \( g(x) dx \) with respect to \( x \)
integration
Integration is a fundamental operation in calculus used to find functions such as antiderivatives. In the context of differential equations, it’s a crucial step in solving both linear and non-linear equations. Once the differential equation is simplified appropriately using methods like separation of variables or integrating factors, integrating both sides can lead to the solution.
In practice, integration involves finding a function whose derivative matches the integrand. Common techniques include:
In practice, integration involves finding a function whose derivative matches the integrand. Common techniques include:
- Basic antiderivatives (e.g., \( \frac{1}{x} \) integrals)
- Integration by parts
- Partial fraction decomposition
- Using substitution (like \( u \)-substitution)