November 5, 2018

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

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