Solving inexact ODE

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

Does someone please know how they combined the two equations for H to get the finial equation $H(x,y) = .... = c$?

Thanks!

 

[Orodruin]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 345763
View attachment 345764
Does someone please know how they combined the two equations for H to get the finial equation $H(x,y) = .... = c$?

Thanks!

Take $H-H$ using both expressions. You find that
$$
0=H-H = \xi_1(y) - \frac{y^2}{2} - \xi_2(x)
$$
Moving the $x$-dependent term to the LHS
$$
\xi_2(x) = \xi_1(y) - \frac{y^2}{2}
$$
The LHS depends only on $x$, but the RHS does not depend on $x$ so both sides must be equal to the same constant $-c$. It follows that
$$\xi_2(x) = c$$
and so
$$-xy +\frac{y^2}{2} + x^2y = c$$

 

[Mark44]

Here's another way to explain how they came up with H(x, y).

In the first screen shot of the OP, the author found that $H(x, y) = x^2 - xy + \xi_1(y)$ and that $H(x, y) = -xy + \frac{y^2}2 + x^2y + \xi_2(x)$. The $\xi_i$ functions are functions of a single variable only.

Since the 2nd version of H(x, y) contains a term in y alone, namely $\frac{y^2}2$, this means that $\xi_1(y) = \frac{y^2}2$. Also, since the 1st version of H contains no term in x alone, this means that $\xi_2(x) = 0$. After all, both versions of H(x, y) must be the same.

Hence $H(x, y) = x^2y - xy + \frac{y^2}2$

The differential equation that was derived from the original pair of equations (that involved t) is $(2xy - y)dx +(-x + y + x^2)dy = 0$. This can be seen as $\frac{\partial H(x, y)}{\partial x}dx + \frac{\partial H(x, y)}{\partial y} dy = 0$.

The LHS of the last equation is the total differential of H(x, y). Since the total differential is zero, it must be true that H(x, y) = c, a constant.

BTW, the thread title is misleading, since the equation you're dealing with is exact, not inexact.

 

[member 731016]

Thank you for your replies @Orodruin and @Mark44!

Sorry there is a typo in the title. It should be a exact equation.

Thanks!