First Order ODE/(x exp y + y - x^2) dy = (2 x y - exp y - x) dx

From ProofWiki
Jump to navigation Jump to search

Theorem

The first order ordinary differential equation:

$(1): \quad \paren {x e^y + y - x^2} \rd y = \paren {2 x y - e^y - x} \rd x$

is an exact differential equation with solution:

$2 x e^y + x^2 + y^2 - 2 x^2 y = C$


Proof

Let $(1)$ be expressed in the form:

$\paren {e^y + x - 2 x y} \rd x + \paren {x e^y + y - x^2} rd y = 0$


Let:

$\map M {x, y} = e^y + x - 2 x y$
$\map N {x, y} = x e^y + y - x^2$

Then:

\(\ds \dfrac {\partial M} {\partial y}\) \(=\) \(\ds e^y - 2 x\)
\(\ds \dfrac {\partial N} {\partial x}\) \(=\) \(\ds e^y - 2 x\)


Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.


By Solution to Exact Differential Equation, the solution to $(1)$ is:

$\map f {x, y} = C$

where:

\(\ds \dfrac {\partial f} {\partial x}\) \(=\) \(\ds \map M {x, y}\)
\(\ds \dfrac {\partial f} {\partial y}\) \(=\) \(\ds \map N {x, y}\)


Hence:

\(\ds f\) \(=\) \(\ds \map M {x, y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds \int \paren {e^y + x - 2 x y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds x e^y + \frac {x^2} 2 - x^2 y + \map g y\)

and:

\(\ds f\) \(=\) \(\ds \int \map N {x, y} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds \int \paren {x e^y + y - x^2} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds x e^y + \frac {y^2} 2 - x^2 y + \map h x\)

Thus:

$\map f {x, y} = x e^y + \dfrac {x^2 + y^2} 2 - x^2 y$

and by Solution to Exact Differential Equation, the solution to $(1)$ is:

$2 x e^y + x^2 + y^2 - 2 x^2 y = C$

after multiplying by $2$ to clear the fraction.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=First_Order_ODE/(x_exp_y_%2B_y_-_x%5E2)_dy_%3D_(2_x_y_-_exp_y_-_x)_dx&oldid=601054"