Question
upstudy study bank question image url

\( ( 6 x ^ { 2 } + 2 y ^ { 2 } ) d x - ( x ^ { 2 } + 4 x y ) d y = 0 \)

Ask by Schneider Hall. in Nigeria
Jan 27,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The general solution is \( 6x - y - \frac{2y^{2}}{x} = C \).

Solution

To solve the differential equation: \[ (6x^{2} + 2y^{2})\,dx - (x^{2} + 4xy)\,dy = 0 \] we can follow these steps: ### 1. **Check for Exactness** First, write the equation in the standard form \( M(x, y)\,dx + N(x, y)\,dy = 0 \): \[ M = 6x^{2} + 2y^{2}, \quad N = -(x^{2} + 4xy) \] Compute the partial derivatives: \[ \frac{\partial M}{\partial y} = 4y, \quad \frac{\partial N}{\partial x} = - (2x + 4y) \] Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is **not exact**. ### 2. **Find an Integrating Factor** To make the equation exact, we look for an integrating factor \( \mu(x) \) that depends only on \( x \). Calculate: \[ \frac{M_y - N_x}{N} = \frac{4y - (-2x - 4y)}{-(x^{2} + 4xy)} = \frac{2x + 8y}{-x^{2} - 4xy} = \frac{-2(x + 4y)}{x(x + 4y)} = \frac{-2}{x} \] Since this expression depends only on \( x \), the integrating factor is: \[ \mu(x) = e^{\int -\frac{2}{x} \, dx} = e^{-2 \ln x} = x^{-2} \] ### 3. **Multiply Through by the Integrating Factor** Multiply the entire differential equation by \( \mu(x) = x^{-2} \): \[ \left(6x^{2} + 2y^{2}\right)x^{-2} dx - \left(x^{2} + 4xy\right)x^{-2} dy = 0 \] Simplify: \[ (6 + 2\frac{y^{2}}{x^{2}})\,dx - \left(1 + 4\frac{y}{x}\right)\,dy = 0 \] Now, the equation is **exact**. ### 4. **Find the Potential Function** Find a function \( \psi(x, y) \) such that: \[ \frac{\partial \psi}{\partial x} = 6 + 2\frac{y^{2}}{x^{2}}, \quad \frac{\partial \psi}{\partial y} = -1 - 4\frac{y}{x} \] Integrate \( \frac{\partial \psi}{\partial x} \) with respect to \( x \): \[ \psi(x, y) = \int \left(6 + 2\frac{y^{2}}{x^{2}}\right) dx = 6x - 2\frac{y^{2}}{x} + h(y) \] Differentiate \( \psi \) with respect to \( y \): \[ \frac{\partial \psi}{\partial y} = -4\frac{y}{x} + h'(y) \] Set this equal to \( -1 - 4\frac{y}{x} \): \[ -4\frac{y}{x} + h'(y) = -1 - 4\frac{y}{x} \implies h'(y) = -1 \implies h(y) = -y + C \] Thus, the potential function is: \[ \psi(x, y) = 6x - \frac{2y^{2}}{x} - y + C \] ### 5. **General Solution** Set \( \psi(x, y) = \text{constant} \): \[ 6x - \frac{2y^{2}}{x} - y = C \] where \( C \) is an arbitrary constant. ### **Final Answer** After simplifying, a general solution is 6 x – y – (2 y²)⁄x equals a constant. Thus, 6x − y − 2y²/x = C

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

Mind Expander

This equation represents a differential equation that can be viewed in the context of exact equations. To determine if it's exact, we would need to check if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), where \( M = 6x^2 + 2y^2 \) and \( N = -(x^2 + 4xy) \). If they are equal, this means the equation can potentially be solved by finding a function whose differential equals this form, giving insights into the behavior of the system being modeled. In real-world applications, a differential equation like this can model various phenomena, such as motion or force balance in physics and engineering. For example, the terms might represent forces acting on a particle, with \( x \) and \( y \) as spatial coordinates. Understanding these relationships can be crucial for tasks like optimizing design in mechanical systems or predicting the behavior of dynamic systems over time.

Latest Calculus Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy