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

2. $\frac{dy}{dx}=\frac{3{y}^{2}+10x{y}^{2}}{2-6xy-10{x}^{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. $\frac{1}{x}dy-\frac{y}{{x}^{2}}dx=0$$\displaystyle\frac{1}{x}dy-\frac{y}{x^2}dx = 0$

2. $\left(2xy+6x\right)\phantom{\rule{0.167em}{0ex}}dx+\left({x}^{2}+4{y}^{3}\right)\phantom{\rule{0.167em}{0ex}}dy=0$$\displaystyle (2xy + 6x)\,dx + (x^2 + 4y^3)\,dy = 0$

3. $\left(3{x}^{2}+y\mathrm{cos}\left(x\right)\right)\phantom{\rule{0.167em}{0ex}}dx+\left(\mathrm{sin}\left(x\right)-4{y}^{3}\right)\phantom{\rule{0.167em}{0ex}}dy=0$$\displaystyle (3x^2 + y \cos{(x)})\,dx + (\sin{(x)} - 4y^3)\,dy = 0$

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

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

Investigation 06

Solve the following IVPs, if possible:

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

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