First Order ODE/Cosine (x + y) dx = sine (x + y) dx + x sine (x + y) dy

Theorem

The first order ordinary differential equation:

$\map \cos {x + y} \rd x = \map \sin {x + y} \rd x + x \map \sin {x + y} \rd y$

is an exact differential equation with solution:

$x \map \cos {x + y} = C$

Proof

Express it in the form:

$\paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + x \map \sin {x + y} \rd y$

Let:

$\map M {x, y} = x \map \sin {x + y} - \map \cos {x + y}$

$\map N {x, y} = x \map \sin {x + y}$

Then:

\(\ds \frac {\partial M} {\partial y}\)

\(=\)

\(\ds x \map \cos {x + y} + \map \sin {x + y}\)

\(\ds \frac {\partial N} {\partial x}\)

\(=\)

\(\ds x \map \cos {x + y} + \map \sin {x + y}\)

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:

\(\ds f\)

\(=\)

\(\ds \map M {x, y} \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds \int \paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds \map \sin {x + y} - x \map \cos {x + y} - \map \sin {x + y}\)

\(\ds \)

\(=\)

\(\ds -x \map \cos {x + y}\)

and:

\(\ds f\)

\(=\)

\(\ds \map N {x, y} \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds \int x \map \sin {x + y} \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds -x \map \cos {x + y} + \map h x\)

Thus:

$\map f {x, y} = -x \map \cos {x + y}$

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

$-x \map \cos {x + y} = C_1$

or setting $C = -C_1$:

$x \map \cos {x + y} = C$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $15$