Is My Solution to This Exact Differential Equation Correct?

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The discussion revolves around solving an exact differential equation represented by the expression (x cos(y) + x^2 + y) dx = - (x + y^2 - (x^2)/2 sin y) dy. The user integrated both sides and derived the equation x^3 + 6xy + y^3 = 0, questioning the correctness of their calculations and whether an alternative method is needed. Participants confirm that the system is indeed exact, as the partial derivatives of the equations match, suggesting the existence of a function f(x,y) that satisfies the conditions. They recommend reviewing the concept of exact equations from textbooks to better understand the solution process. The conversation highlights the simplicity of solving such equations once the method is grasped.
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Homework Statement
(x cos(y) + x^2 +y ) dx + (x + y^2 - (x^2)/2 sin y ) dy = 0
Relevant Equations
(x cos(y) + x^2 +y ) dx = - (x + y^2 - (x^2)/2 sin y ) dy
(x cos(y) + x2 +y ) dx = - (x + y2 - (x2)/2 sin y ) dy
I integrated both sides
1/2x2cos(y) + 1/3 x3+xy = -xy - 1/3y3+x2cos(y)

Then
I get x3 + 6xy + y3 = 0

Am I doing the calculations correctly?
Do I need to solve it in another way?
 
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You are looking for a function f(x,y) such that <br /> \begin{align*}<br /> \frac{\partial f}{\partial x} &amp;= x \cos(y) + x^2 +y \\<br /> \frac{\partial f}{\partial y} &amp;= x + y^2 -\frac12 x^2 \sin y<br /> \end{align*} The solution is then given implicitly by f(x,y(x))) = C.

Are you familiar with how to solve such a system?
 
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pasmith said:
You are looking for a function f(x,y) such that <br /> \begin{align*}<br /> \frac{\partial f}{\partial x} &amp;= x \cos(y) + x^2 +y \\<br /> \frac{\partial f}{\partial y} &amp;= x + y^2 -\frac12 x^2 \sin y<br /> \end{align*} The solution is then given implicitly by f(x,y(x))) = C.

Are you familiar with how to solve such a system?
Then do I have to solve two different differential equations?
 
the first thing i notice is that if i take the partial of the first equation with respect to y it is equal to the partial of the second equation with respect to x. I think this means the system is "exact". My differential equations are a bit rusty. To solve this system, you may need to review exact equations in yout textbook,
 
mpresic3 said:
the first thing i notice is that if i take the partial of the first equation with respect to y it is equal to the partial of the second equation with respect to x. I think this means the system is "exact". My differential equations are a bit rusty. To solve this system, you may need to review exact equations in yout textbook,
Correct. It implies there is f(x,y) with f_x , f_y equal to the respective equations.
 
This is a class of ODE known as exact equations in this text

Here is the theorem that gives the solution and a corresponding example from the text
Screen Shot 2021-04-20 at 12.33.11 AM.png

Screen Shot 2021-04-20 at 12.33.22 AM.png

once you get the hang of the process it becomes pretty simple actually.
 
docnet said:
This is a class of ODE known as exact equations in this text

Here is the theorem that gives the solution and a corresponding example from the textView attachment 281782
View attachment 281783
once you get the hang of the process it becomes pretty simple actually.
I always wondered if it is a coincidence that these are precisely the Cauchy-Riemann equations.