Determine b such that the equation. for value b find the solution for the differential equation.
Determine b such that the equation. for value b find the solution for the differential equation.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Determine b such that the equation. for value b find the solution for the
![The equation given is:
\[ (xy^2 + bx^2y) + (x + y)x^2 \frac{dy}{dt} = 0 \]
This is a differential equation. In this equation:
- \(x\) and \(y\) are variables.
- \(b\) is a constant.
- \(\frac{dy}{dt}\) represents the derivative of \(y\) with respect to \(t\).
The equation suggests that the given expression is exact. An exact differential equation is of the form \( M(x, y) + N(x, y) \frac{dy}{dx} = 0 \), where there exists a function \( \psi(x, y) \) such that:
\[ \frac{\partial \psi}{\partial x} = M \]
\[ \frac{\partial \psi}{\partial y} = N \]
In this case:
\[ M(x, y) = xy^2 + bx^2y \]
\[ N(x, y) = (x + y)x^2 \]
The term "exact" implies that there exists a potential function \( \psi(x, y) \) whose total differential gives the terms \( M(x, y) \) and \( N(x, y) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc47a605b-1167-4b41-bc40-883b330bf596%2Fba71ffea-9998-4180-8fc8-81d292ab145b%2Fyxo1po_processed.png&w=3840&q=75)
Transcribed Image Text:The equation given is:
\[ (xy^2 + bx^2y) + (x + y)x^2 \frac{dy}{dt} = 0 \]
This is a differential equation. In this equation:
- \(x\) and \(y\) are variables.
- \(b\) is a constant.
- \(\frac{dy}{dt}\) represents the derivative of \(y\) with respect to \(t\).
The equation suggests that the given expression is exact. An exact differential equation is of the form \( M(x, y) + N(x, y) \frac{dy}{dx} = 0 \), where there exists a function \( \psi(x, y) \) such that:
\[ \frac{\partial \psi}{\partial x} = M \]
\[ \frac{\partial \psi}{\partial y} = N \]
In this case:
\[ M(x, y) = xy^2 + bx^2y \]
\[ N(x, y) = (x + y)x^2 \]
The term "exact" implies that there exists a potential function \( \psi(x, y) \) whose total differential gives the terms \( M(x, y) \) and \( N(x, y) \).
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