Consider the DE
\[\frac{d y}{d x}-\frac{2}{x} y=x^3 y^2, \quad x>0\]
(a) Identify the type of DE and solve it.
(b) Briefly state the domain of validity of your solution.
Answer (2)
Identify the given differential equation as a Bernoulli equation.
Apply the substitution v = y − 1 to transform the equation into a linear first-order differential equation.
Find the integrating factor and solve the linear equation for v .
Substitute back to find y and determine the domain of validity: y = 6 C − x 6 6 x 2 for 0"> x > 0 and x = ( 6 C ) 1/6 if 0"> C > 0 .
Explanation
Problem Identification and Objective We are given the differential equation d x d y − x 2 y = x 3 y 2 , with the condition 0"> x > 0 . Our goal is to identify the type of differential equation, solve it, and determine the domain of validity for the solution.
Recognizing Bernoulli Equation and Substitution The given differential equation is a Bernoulli equation. Bernoulli equations have the form d x d y + P ( x ) y = Q ( x ) y n . In our case, P ( x ) = − x 2 , Q ( x ) = x 3 , and n = 2 . To solve this, we use the substitution v = y 1 − n = y 1 − 2 = y − 1 .
Finding the Derivative and Substituting Now, we find the derivative of v with respect to x : d x d v = − y − 2 d x d y , which means d x d y = − y 2 d x d v . Substituting this into the original equation, we get − y 2 d x d v − x 2 y = x 3 y 2 .
Transforming into a Linear Equation Dividing the equation by − y 2 , we obtain d x d v + x 2 y − 1 = − x 3 . Since v = y − 1 , the equation becomes d x d v + x 2 v = − x 3 . This is a linear first-order differential equation in terms of v .
Finding the Integrating Factor To solve the linear equation, we find the integrating factor μ ( x ) = e ∫ P ( x ) d x = e ∫ x 2 d x = e 2 l n x = e l n x 2 = x 2 .
Multiplying by the Integrating Factor Multiplying the linear equation by the integrating factor x 2 , we get x 2 d x d v + 2 xv = − x 5 . The left side is the derivative of x 2 v , so we have d x d ( x 2 v ) = − x 5 .
Integrating Both Sides Integrating both sides with respect to x , we get ∫ d x d ( x 2 v ) d x = ∫ − x 5 d x , which gives x 2 v = − 6 x 6 + C , where C is the constant of integration.
Solving for v and Substituting Back Solving for v , we have v = − 6 x 4 + x 2 C . Substituting back v = y − 1 , we get y 1 = − 6 x 4 + x 2 C = 6 x 2 6 C − x 6 .
Solving for y Finally, solving for y , we get y = 6 C − x 6 6 x 2 .
Determining the Domain of Validity Now, we determine the domain of validity. Since 0"> x > 0 , we need to ensure that 6 C − x 6 = 0 , so x 6 = 6 C . If 0"> C > 0 , then x = ( 6 C ) 1/6 . If C ≤ 0 , then 6 C − x 6 < 0 for all 0"> x > 0 , so y < 0 . The domain of validity is 0"> x > 0 , excluding x = ( 6 C ) 1/6 if 0"> C > 0 .
Final Solution and Domain of Validity Therefore, the solution to the differential equation is y = 6 C − x 6 6 x 2 , and the domain of validity is 0"> x > 0 , x = ( 6 C ) 1/6 if 0"> C > 0 .
Examples
Bernoulli equations appear in various contexts, such as fluid dynamics and population dynamics. For instance, in fluid dynamics, they can model the flow of a fluid through a pipe with varying cross-sectional area. In population dynamics, they can describe the growth of a population subject to certain constraints or harvesting. Understanding how to solve these equations allows engineers and scientists to analyze and predict the behavior of these systems.
The given differential equation is a Bernoulli equation that can be solved by substitution. The solution is y = 6 C − x 6 6 x 2 , valid for 0"> x > 0 and x = ( 6 C ) 1/6 if 0"> C > 0 .
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