Exact Equations

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 2.4 – Exact Equations

Expected Educational Results

• Objective 6–1: I can identify the exact differential form.

• Objective 6–2: I can determine if an ODE is an exact equation.

• Objective 6–3: I can solve exact equations.

Exact Equations

Investigation 04

Determine if the following DEs are exact equations:

1. ${\displaystyle \frac{dy}{dx}=-y}$$\displaystyle \frac{dy}{dx}=-y$

2. ${\displaystyle \frac{dy}{dx}=\frac{3y^2+10x{y}^2}{2-6xy-10x^{2}y}}$$\displaystyle \frac{dy}{dx}=\frac{3y^2 + 10xy^2}{2 - 6xy - 10x^2y}$

Investigation 05

Determine if the following DEs are exact equations. Solve, if possible:

1. ${\displaystyle \frac{1}{x}dy-\frac{y}{x^2}dx=0}$$\displaystyle\frac{1}{x}dy-\frac{y}{x^2}dx = 0$

2. ${\displaystyle (2xy+6x)dx+(x^2+4y^3)dy=0}$$\displaystyle (2xy + 6x)\,dx + (x^2 + 4y^3)\,dy = 0$

3. ${\displaystyle (3x^2+y\cos (x))dx+(\sin (x)-4y^3)dy=0}$$\displaystyle (3x^2 + y \cos{(x)})\,dx + (\sin{(x)} - 4y^3)\,dy = 0$

4. ${\displaystyle \frac{dy}{dx}=\frac{2x^3-3x^{2}y+y^3}{6x^{2}y-2x^3-3x{y}^2}}$$\displaystyle \frac{dy}{dx} = \frac{2x^3 - 3x^2 y + y^3}{6x^2y - 2x^3 - 3xy^2}$

5. ${\displaystyle \frac{dy}{dx}=\frac{3x(2-xy)}{x^3+2y}}$$\displaystyle \frac{dy}{dx} = \frac{3x(2 - xy)}{x^3 + 2y}$

Investigation 06

Solve the following IVPs, if possible:

1. ${\displaystyle \frac{dy}{dx}=-\frac{y{e}^{xy}}{x{e}^{xy}+\sin (y)}}$$\displaystyle \frac{dy}{dx} = -\frac{ye^{xy}}{xe^{xy} + \sin{(y)}}$, $y(1)=\pi$$y(1) = \pi$

2. ${\displaystyle (\cos (y)+y\cos (x))dx+(\sin (x)-x\sin (y))dy=0}$$\displaystyle (\cos{(y)} + y \cos{(x)}) dx + (\sin{(x)} - x \sin{(y)}) dy = 0$, $y(0)=-1$$y(0) = -1$

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Last Modified: Wednesday, 2 September 2020 07:42 EDT