First Order ODE/(y^2 exp x y + cosine x) dx + (exp x y + x y exp x y) dy = 0
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Theorem
The first order ordinary differential equation:
- $(1): \quad \paren {y^2 e^{x y} + \cos x} \rd x + \paren {e^{x y} + x y e^{x y} } \rd y = 0$
is an exact differential equation with solution:
- $y e^{x y} + \sin x = C$
Proof
Let:
- $\map M {x, y} = y^2 e^{x y} + \cos x$
- $\map N {x, y} = e^{x y} + x y e^{x y}$
Then:
\(\ds \dfrac {\partial M} {\partial y}\) | \(=\) | \(\ds 2 y e^{x y} + x y^2 e^{x y}\) | ||||||||||||
\(\ds \dfrac {\partial N} {\partial x}\) | \(=\) | \(\ds y e^{x y} + x y^2 e^{x y} + y e^{x y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 y e^{x y} + x y^2 e^{x y}\) |
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:
\(\ds f\) | \(=\) | \(\ds \map M {x, y} \rd x + \map g y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {y^2 e^{x y} + \cos x} \rd x + \map g y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds y e^{x y} + \sin x + \map g y\) |
and:
\(\ds f\) | \(=\) | \(\ds \int \map N {x, y} \rd y + \map h x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {e^{x y} + x y e^{x y} } \rd y + \map h x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{x y} } x + y e^{x y} - \frac {e^{x y} } x + \map h x\) | Primitive of $x e^{a x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds y e^{x y} + \map h x\) |
Thus:
- $\map f {x, y} = y e^{x y} + \sin x$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
- $y e^{x y} + \sin x = C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $17$