Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 2: Problem 21
In Problems 21 and 22 solve the given Bernoulli equation subject to the indicated initial condition. \(x^{2} \frac{d y}{d x}-2 x y=3 y^{4}, \quad y(1)=\frac{1}{2}\)
Short Answer
Expert verified
The solution is \( y(x) = \left( \frac{9}{7x} + Cx^6 \right)^{-1/3} \), where \( C \) is determined by \( y(1) = \frac{1}{2} \).
Step by step solution
01
Identify Bernoulli's Equation
The given differential equation is \( x^{2} \frac{dy}{dx} - 2xy = 3y^4 \). A Bernoulli's equation has the form \( \frac{dy}{dx} + P(x)y = Q(x)y^n \). Rewriting the given equation: \( \frac{dy}{dx} = \frac{2xy}{x^2} + \frac{3y^4}{x^2} \). This matches the Bernoulli form with \( P(x) = -\frac{2}{x} \), \( Q(x) = \frac{3}{x^2} \), and \( n = 4 \).
02
Divide Through by \(y^4\)
Divide every term in the equation by \( y^4 \) to rewrite it as: \( y^{-4} \frac{dy}{dx} - \frac{2}{x} y^{-3} = \frac{3}{x^2} \). This prepares the equation for a substitution to make it linear.
03
Use Substitution to Transform Equation
Let \( v = y^{-3} \), so \( \frac{dv}{dx} = -3y^{-4} \frac{dy}{dx} \). The differential equation becomes \( -\frac{1}{3} \frac{dv}{dx} + \frac{2}{x}v = \frac{3}{x^2} \). Rearranging gives \( \frac{dv}{dx} - \frac{6}{x}v = -\frac{9}{x^2} \).
04
Find the Integrating Factor
The equation \( \frac{dv}{dx} - \frac{6}{x} v = -\frac{9}{x^2} \) is linear in \( v \). To solve this using an integrating factor, calculate \( \mu(x) = e^{\int -\frac{6}{x} dx} = e^{-6 \ln x} = x^{-6} \).
05
Apply Integrating Factor to Solve for \(v(x)\)
Multiply every term in the differential equation by \( x^{-6} \): \( x^{-6} \frac{dv}{dx} - 6x^{-7}v = -\frac{9}{x^8} \). This simplifies to \( \frac{d}{dx}(x^{-6}v) = -\frac{9}{x^8} \). Integrate both sides: \( x^{-6}v = \int -\frac{9}{x^8} dx = \frac{9}{7x^7} + C \).
06
Solve for \(v(x)\)
Solve for \( v(x) \): \( v(x) = x^{6} \left( \frac{9}{7x^7} + C \right) = \frac{9}{7x} + Cx^{6} \).
07
Convert Back to \(y(x)\)
Since \( v = y^{-3} \), we have \( y(x) = v^{-1/3} = \left( \frac{9}{7x} + Cx^{6} \right)^{-1/3} \).
08
Use Initial Condition to Find \(C\)
Apply the initial condition \( y(1) = \frac{1}{2} \). Substitute \( x = 1 \) and \( y = \frac{1}{2} \) into \( \left( \frac{9}{7} + C \right)^{-1/3} = \frac{1}{2} \). Solving gives \( C = \frac{119}{8} - \frac{9}{7} \). Calculate to find \( C \).
09
Write the Final Solution
Substitute \( C \) back into \( y(x) = \left( \frac{9}{7x} + \left( \frac{119}{8} - \frac{9}{7} \right)x^{6} \right)^{-1/3} \). This is the solution to the differential equation with the given initial condition.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Condition
The initial condition in a differential equation provides a specific solution among potentially infinite others. It means that we are given the value of the solution at a particular point. In our example, we have the condition \( y(1) = \frac{1}{2} \), which gives us a value of the function \( y \) at \( x = 1 \). This information allows us to find the constant \( C \) after solving the differential equation. Without it, we would only arrive at a general solution containing an arbitrary constant. The initial condition helps tailor the solution to a specific scenario.
Integrating Factor
An integrating factor helps solve linear differential equations. It's essentially a function we multiply with every term to make the equation easier to integrate. For the differential equation in our problem, \( \frac{dv}{dx} - \frac{6}{x}v = -\frac{9}{x^2} \), we find the integrating factor \( \mu(x) = x^{-6} \). The computation involves exponentiating the integral of the coefficient of \( v \), i.e., \( -\frac{6}{x} \). This step converts the left side of the equation into a perfect derivative, reducible through integration. The integrating factor simplifies the equation, making it solvable by direct integration.
Substitution Method
Substitution is a powerful technique to transform differential equations into simpler forms. In our case, the original Bernoulli differential equation isn't linear. We use the substitution \( v = y^{-3} \) to convert it into a linear form. By letting \( v = y^{-3} \), we introduce a new variable that absorbs the non-linear part. The task then becomes solving the new differential equation with respect to \( v \). It turns the nonlinear Bernoulli equation into a standard linear problem, which can be tackled with the integrating factor method. Substitution is crucial for handling equations unsolvable by standard methods.
Linear Differential Equation
A linear differential equation is an equation involving a function and its derivatives, with no products of the function and its derivatives. It's often written in standard form \( \frac{dy}{dx} + P(x)y = Q(x) \).In our problem, the original Bernoulli equation is not linear because of the term \( y^4 \). Using the substitution method, the form \( \frac{dv}{dx} - \frac{6}{x}v = -\frac{9}{x^2} \) becomes linear with respect to \( v \). Linear equations are simpler to solve as they allow methods like integrating factors and direct integration. Understanding how to transform and handle these equations is crucial in differential calculus.
