Solution to Linear First Order Ordinary Differential Equation/Proof 1

Proof
Consider the first order ordinary differential equation:


 * $M \left({x, y}\right) + N \left({x, y}\right) \dfrac {\mathrm d y} {\mathrm d x} = 0$

We can put our equation:
 * $(1) \quad \dfrac {\mathrm d y}{\mathrm d x} + P \left({x}\right) y = Q \left({x}\right)$

into this format by identifying:
 * $M \left({x, y}\right) \equiv P \left({x}\right) y - Q \left({x}\right), N \left({x, y}\right) \equiv 1$

We see that:
 * $\dfrac {\partial M} {\partial y} - \dfrac {\partial N}{\partial x} = P \left({x}\right)$

and hence:
 * $P \left({x}\right) = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N}{\partial x}} N$

is a function of $x$ only.

It immediately follows from Integrating Factor for First Order ODE that:
 * $e^{\int P \left({x}\right) dx}$

is an integrating factor for $(1)$.

So, multiplying $(1)$ by this factor:
 * $e^{\int P \left({x}\right) \ \mathrm d x} \dfrac {\mathrm d y} {\mathrm d x} + e^{\int P \left({x}\right) \ \mathrm d x} P \left({x}\right) y = e^{\int P \left({x}\right) \ \mathrm d x} Q \left({x}\right)$

The result follows by an application of Solution to Exact Differential Equation.