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Question 4 Solve the following differential equation. $x^2 y dx + y \left(x^3 + \frac{\sin y}{e^{3y}}\right) dy = 0$ $\circ \phi(x, y) = \frac{x^3 e^{3y}}{3} + \sin y$ $\circ \phi(x, y) = \frac{x^3 e^{-3y}}{3} - \cos y$ $\circ$ None of the given choices $\circ \phi(x, y) = \frac{x^3 e^{3y}}{3} - \cos y$ $\circ \phi(x, y) = \frac{x^2 e^{3y}}{3} + \cos y$

          Question 4
Solve the following differential equation.
$x^2 y dx + y \left(x^3 + \frac{\sin y}{e^{3y}}\right) dy = 0$
$\circ \phi(x, y) = \frac{x^3 e^{3y}}{3} + \sin y$
$\circ \phi(x, y) = \frac{x^3 e^{-3y}}{3} - \cos y$
$\circ$ None of the given choices
$\circ \phi(x, y) = \frac{x^3 e^{3y}}{3} - \cos y$
$\circ \phi(x, y) = \frac{x^2 e^{3y}}{3} + \cos y$

Question 4
Solve the following differential equation.
x^2 y dx + y (x^3 + (sin y)/(e^3y)) dy = 0
∘ϕ(x, y) = (x^3 e^3y)/(3) + sin y
∘ϕ(x, y) = (x^3 e^-3y)/(3) - cos y
∘ None of the given choices
∘ϕ(x, y) = (x^3 e^3y)/(3) - cos y
∘ϕ(x, y) = (x^2 e^3y)/(3) + cos y

Added by Thomas C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Jacob Mclean

04/10/2022

Step 1

Step 1: The given differential equation is: $$x^2ydx + y(x^3 + \frac{sin y}{e^{3y}})dy = 0$$ Show more…

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Question 4 Solve the following differential equation. $$x^2ydx + y(x^3 + \frac{sin y}{e^{3y}})dy = 0$$ $$ \phi(x, y) = \frac{x^3e^{3y}}{3} + sin y$$ $$ \phi(x, y) = \frac{x^3e^{-3y}}{3} - cos y$$ None of the given choices $$ \phi(x, y) = \frac{x^3e^{3y}}{3} - cos y$$ $$ \phi(x, y) = \frac{x^2e^{3y}}{3} + cos y$$