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Magazine Chapter 2: Problem 6
Classify each differential equation as separable, exact, linear, homogeneous, or Bernoulli. Some equations may be more than one kind. Do not solve the equations. $$\frac{d y}{d x}=5 y+y^{2}$$
Short Answer
Expert verified
The equation is separable and Bernoulli.
Step by step solution
01
Identify the Type - Separable Equation
A differential equation is separable if it can be expressed as the product of a function of \( y \) and a function of \( x \). The given equation \( \frac{dy}{dx} = 5y + y^2 \) can be rewritten as \( \frac{dy}{(5y + y^2)} = dx \), which is separable since it helps separate variables with \( y \) terms on one side and \( x \) terms on the other.
02
Check for Linearity
A linear differential equation in standard form is \( \frac{dy}{dx} + P(x)y = Q(x) \). Our equation \( \frac{dy}{dx} = 5y + y^2 \) contains a term \( y^2 \), indicating it is not linear because linear equations cannot contain \( y^2 \) or higher powers of \( y \).
03
Determine if it's Exact
A differential equation \( M(x, y)dx + N(x, y)dy = 0 \) is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Since our equation is not in this form, and because it does not satisfy the conditions for exactness, it is not exact.
04
Identify if it's Homogeneous
A differential equation is homogeneous if it can be expressed in such a way that all terms are of the same degree. Multiplying each term in \( \frac{dy}{dx} = 5y + y^2 \) by \( x^0y^2 \), shows components of unequal degree, hence the equation is not homogeneous.
05
Determine if it's Bernoulli
A Bernoulli equation is of the form \( \frac{dy}{dx} + P(x)y = Q(x)y^n \). Here, with \( \frac{dy}{dx} = 5y + y^2 \), rearranging gives \( \frac{dy}{dx} - 5y = y^2 \) which is in the form of a Bernoulli equation with \( n = 2 \), \( P(x) = -5 \), and \( Q(x) = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separable Equations
A separable equation lets us split the equation so each side involves only one variable. This means, if we have a differential equation in the form \( \frac{dy}{dx} = g(y)h(x) \), we can rewrite it to separate the variables, giving us: \[ \frac{dy}{g(y)} = h(x)dx \] This makes it easier to solve by integrating both sides separately.
- Step-by-step: Isolate terms with \(y\) on one side and \(x\) on the other.
- Integrate both sides.
- Use initial conditions if available, to find a specific solution.
Exact Differential Equations
Exact differential equations involve a specific relationship between the components of the equation. They take the form: \[ M(x, y)dx + N(x, y)dy = 0 \] The equation is exact if the mixed partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) are equal.
- How to check: Calculate the partial derivatives and check for equality.
- Solving: If exact, integrate \( M(x, y) \) with respect to \( x \) and \( N(x, y) \) with respect to \( y \), ensuring consistency.
Linear Differential Equations
Linear equations conform to the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] They should not have terms like \( y^2 \) or any higher powers of \( y \), as such terms make the equation nonlinear.
- Recognition: Check if all terms of \( y \) have a degree of one.
- Solving Method: Often involves integrating factor method, which simplifies solving.
Homogeneous Equations
Homogeneous differential equations are characterized by variable terms all having the same degree. These can be solved using a substitution method or by recognizing the degree. For an equation \( \frac{dy}{dx} = f(x,y) \) to be homogeneous, every term \( f(ax, ay) = a^n f(x, y) \).
- Identifying: Rewrite and check if each term is of the same degree.
- Resolution technique: Employ substitutions like \( v = \frac{y}{x} \).
Bernoulli Differential Equations
These equations have a form that combines characteristics of linear and nonlinear equations, expressed as: \[ \frac{dy}{dx} + P(x)y = Q(x)y^n \] Bernoulli's equation integrates a linear setup with a nonlinear term \( y^n \).
- Solution process: Often by a substitution \( v = y^{1-n} \), turning the equation into a linear form.
