Is My Solution to This Exact Differential Equation Correct?

potatocake
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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.
 
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