How do I solve this differential equation?

[potatocake]

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?

 

[pasmith]

You are looking for a function $f(x,y)$ such that $$\begin{align*} \frac{\partial f}{\partial x} &= x \cos(y) + x^2 +y \\ \frac{\partial f}{\partial y} &= x + y^2 -\frac12 x^2 \sin y \end{align*}$$ The solution is then given implicitly by $f(x,y(x))) = C$.

Are you familiar with how to solve such a system?

 

[potatocake]

You are looking for a function $f(x,y)$ such that $$\begin{align*} \frac{\partial f}{\partial x} &= x \cos(y) + x^2 +y \\ \frac{\partial f}{\partial y} &= x + y^2 -\frac12 x^2 \sin y \end{align*}$$ 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?

 

[mpresic3]

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,

 

[WWGD]

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.

 

[docnet]

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

once you get the hang of the process it becomes pretty simple actually.

 

[WWGD]

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


View 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.