[Math] Exact Differential Equation

Form

$$ M(x, y) + N(x, y)y' = 0 $$

Test for Exactness


$$ \dfrac{ \partial N }{ \partial x } = \dfrac{ \partial M }{ \partial y } $$

Suppose M, N, ∂N/∂x and ∂M/∂y are continuous for all (x, y) in a rectangle R in the plane having sides parallel to the axes. Then M + Ny' = 0 is exact on R if and only if $$ \dfrac{ \partial N }{ \partial x } = \dfrac{ \partial M }{ \partial y } $$ for (x, y) in R.

Integrating Factor

$$ \dfrac{\partial}{\partial x}(\mu N) = \dfrac{\partial}{\partial y}(\mu M) $$

Examples

Ex. 1

$$
( y \cos x + 2xe^y ) + ( \sin x + x^2 e^y - 1 ) y' = 0
$$

$$
\begin{split}

M(x, y) &= y \cos x + 2xe^y \\
N(x, y) &= \sin x + x^2 e^y - 1 ^y \\
\\
M_y(x, y) &= \cos x + 2 x e^y \\
N_x(x, y) &= \cos x + 2x e^y \\

\end{split}
$$

$$
\begin{split}

&\because M_y(x, y) = N_x(x, y) \\
&\therefore ( y \cos x + 2xe^y ) + ( \sin x + x^2 e^y - 1 ) y' = 0 \text{ is an exact equation}

\end{split}
$$

$$
\begin{split}

F_x(x, y) = M(x, y) &= y \cos x + 2 x e^y \\
F_y(x, y) = N(x, y) &= \sin x + x^2 e^y - 1 \\
F(x, y) &= y \sin x + x^2 e^y + h(y) \text{  from } M(x, y) \\
F_y(x, y) &= \sin x + x^2 e^y + h'(y) \text{  from } F(x, y) \\
&= \sin x + x^2 e^y - 1 \\
&\implies h'(y) = -1 \\
&\implies h(y) = -y

\end{split}
$$

$$
F(x, y) = y \sin x + x^2 e^y - y \\
\text{General Solution: } y \sin x + x^2 e^y - y = C
$$

Ex. 2

$$ ( 2 x + y + 1 ) dx + ( 2 y + x + 1 ) dy = 0 $$

$$
\begin{split}

M(x, y) &= 2 x + y + 1 \\
N(x, y) &= 2 y + x + 1 \\
\\
M_y(x, y) &= 1 \\
N_x(x, y) &= 1 \\

\end{split}
$$

$$
\begin{split}

&\because M_y(x, y) = N_x(x, y) \\
&\therefore ( 2 x + y + 1 ) dx + ( 2 y + x + 1 ) dy = 0 \text{ is an exact equation}

\end{split}
$$

$$
\begin{split}

F_x(x, y) = M(x, y) &= 2 x + y + 1 \\
F_y(x, y) = N(x, y) &= 2 y + x + 1 \\
F(x, y) &= x^2 + x y + x + g(y) \text{  from } M(x, y) \\
F(x, y) &= y^2 + x y + y + h(y) \text{  from } N(x, y) \\
&\implies g(y) = y^2 + y \\
&\implies h(y) = x^2 + x

\end{split}
$$

$$
\begin{split}

F(x, y) &= C \\
\text{General Solution: } x^2 + y^2 + x y + x + y &= C

\end{split}
$$

Ex. 3

$$
\begin{split}

( 3 x y + y^2 ) + ( x^2 + x y ) &= 0 \\
\\
M_y = 3 x + 2 y &\neq N_x = 2 x + y \\
\\
\mu(x)(3 x y + y^2) + \mu(x)(x^2 + x y) y' &= 0 \\
\mu(x)(3 x + 2 y) &= \mu'(x)(x^2 + x y) + \mu(x)(2 x + y) \\
\mu(x)(3 x + 2 y) - \mu(x)(2 x + y) &= \mu'(x)(x^2 + x y ) \\
\mu(x)(3 x + 2 y - 2 x - y) &= \mu'(x)(x^2 + x y) \\
\mu(x)(x + y) &= \mu'(x) x (x + y) \\
\mu(x) &= \mu'(x) x \\
\mu(x) &= \dfrac{du}{dx} x \\
\dfrac{\mu(x)}{x} &= \dfrac{du}{dx} \\
\dfrac{1}{x} &= \dfrac{1}{u} \dfrac{du}{dx} \\
\dfrac{1}{x} dx &= \dfrac{1}{u} du \\
\mu(x) &= x \\
\\
\mu(x) (3 x y + y^2) + \mu(x) (x^2 + x y) y' &= 0 \\
x (3 x y + y^2) + x (x^2 + x y) y' &= 0 \\
(3 x^2 y + x y^2) + (x^3 + x^2 y) y' &= 0 \\
\\
3 x^2 + 2 x y &= 3 x^2 + 2 x y \\
\psi_x &= 3 x^2 y + x y^2 \\
\psi &= x^3 y + \dfrac{1}{2}x^2 y^2 + h(y) \\
\psi_y &= x^3 + x^2 y + h'(y) = x^3 + x^2 y \\
& \implies h'(y) = 0, h(y) = C \\
\\
x^3 + \frac{1}{2} x^2 y^2 &= C

\end{split}
$$

Reference

https://www.youtube.com/watch?v=bwASJWS8ltM
https://www.youtube.com/watch?v=u1takDsKQfk
https://www.youtube.com/watch?v=j511hg7Hlbg
https://www.youtube.com/watch?v=0NyeDUhKwBE

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