Lesson 8 (09/11/2026)

Lesson 8 (09/11/2026)

Exact Equations (1.6)

Warmup: \( y \) is implicitly defined as a function of \( x \)

\[ \phi(x, y) = \sin(xy) + x + y^3 = 0, \quad \text{find } y' \]

take \( \frac{d}{dx} \) on both sides, Apply Chain Rule

\[ \frac{d}{dx} \left[ \sin(xy) + x + y^3 \right] = \frac{d}{dx} 0 = 0 \] \[ \cos(xy) \cdot \frac{d}{dx} [xy] + 1 + 3y^2 \cdot \frac{dy}{dx} = 0 \] \[ \cos(xy) \left[ y + x \frac{dy}{dx} \right] + 1 + 3y^2 \left( \frac{dy}{dx} \right) = 0 \] \[ \left[ x \cos(xy) + 3y^2 \right] \frac{dy}{dx} + [1 + y \cos(xy)] = 0 \] \[ \frac{dy}{dx} = -\frac{[1 + y \cos(xy)]}{[x \cos(xy) + 3y^2]} = -\frac{\phi_x}{\phi_y} \]

Side Calculations:

\[ \phi = \sin(xy) + x + y^3 \] \[ \phi_x = y \cos(xy) + 1 \] \[ \phi_y = x \cos(xy) + 3y^2 \]

Differential equation:

\[ \frac{dy}{dx} = -\frac{\phi_x}{\phi_y} \]

or

\[ \phi_x + \phi_y \frac{dy}{dx} = 0 \]

has \( \phi(x, y) = c \) is solution given implicitly

Today:

Given \( M(x, y) + N(x, y)\frac{dy}{dx} = 0 \)

• What are conditions on \( M, N \) so that you find a solution like above
• How do we find the solution.

Calc 3:

\[ M(x,y) + N(x,y)\frac{dy}{dx} = 0 \]

is there a function \( \phi(x,y) \) such that

\[ \phi_x = M(x,y), \quad \phi_y = N(x,y) \]

1. Mixed partials are the same

\[ \phi_{xy} = \phi_{yx} \]

\[ \left. \begin{aligned} \phi_x = M &\implies \phi_{xy} = M_y \\ \phi_y = N &\implies \phi_{yx} = N_x \end{aligned} \right\} \text{want } M_y = N_x \]

2. \( \vec{F} = M\vec{i} + N\vec{j} \) is conservative if

• \( \exists \phi \) such that \( \vec{F} = \nabla\phi = \phi_x\vec{i} + \phi_y\vec{j} \)
• \( \text{2D curl } \vec{F} = M_y - N_x = 0 \)

Exact Equations:

Differential equation:

\[ M(x, y) + N(x, y)\frac{dy}{dx} = 0 \]

OR

Differential form

\[ M(x, y)\,dx + N(x, y)\,dy = 0 \]

is Exact if there exist a \( \phi(x, y) \) such that

\[ M(x, y) = \phi_x \quad \text{and} \quad N(x, y) = \phi_y \]

if such a \( \phi(x, y) \) exists then

\[ \phi(x, y) = C \quad \text{is the General solution,} \]

only possible when

\[ M_y = N_x \]

Exactness condition

Which of the following are Exact Equations?

1. \( (\sin x + y)\,dx + (x + e^y)\,dy = 0 \)

\[ \left. \begin{aligned} M &= \sin x + y \implies M_y = 1 \\ N &= x + e^y \implies N_x = 1 \end{aligned} \right\} M_y = N_x \quad \text{exact} \]

2. \( (\sin x + xy)\,dx + (e^y + xy)\,dy = 0 \)

\[ \left. \begin{aligned} M &= \sin x + xy \implies M_y = x \\ N &= e^y + xy \implies N_x = y \end{aligned} \right\} M_y \neq N_x \quad \text{Not exact} \]

3. \( (3y + 6x)\,dx + (6y + 3x)\,dy = 0 \)

\[ \left. \begin{aligned} M &= 3y + 6x \implies M_y = 3 \\ N &= 6y + 3x \implies N_x = 3 \end{aligned} \right\} M_y = N_x \quad \text{exact} \]

eg: \( (\sin x + y)dx + (e^y + x)dy = 0 \) is exact

How do we find solution \( \phi(x,y) \).

• find \( \phi \) such that \[ 1) \phi_x = M = \sin x + y \] \[ 2) \phi_y = N = e^y + x \]

integrate w.r.t \( x \)

\[ \phi(x,y) = \int \sin x + y \, dx + C(y) \]

Constant w.r.t \( x \) so can be a function of \( y \), \( \frac{\partial}{\partial x}(C(y)) = 0 \).

\[ \phi(x,y) = -\cos x + xy + C(y) \]

\[ \frac{\partial \phi}{\partial y} = e^y + x \]

plug in \( \phi(x, y) = -\cos x + xy + C(y) \)

\[ \frac{\partial}{\partial y} [-\cos x + xy + C(y)] = e^y + x \] \[ 0 + x + \frac{dC}{dy} = e^y + x \] \[ \frac{dC}{dy} = e^y \]

integrate w.r.t y on both side

\[ C(y) = e^y + k \]

\[ \phi(x, y) = -\cos x + xy + e^y + K \]

Satisfies

\[ \phi_x = \sin x + y \]

\[ \phi_y = e^y + x \]

\( \therefore \phi(x, y) = -\cos x + xy + e^y + k = 0 \) is the General solution to

\[ \phi_x + \phi_y \frac{dy}{dx} = 0 \]

OR

\[ [\sin x + y] + [e^y + x] \frac{dy}{dx} = 0 \]

Example:

\[ (3y+6x) + (6y+3x)\frac{dy}{dx} = 0 \]

is Exact [Already Verified].

to find solution \( \Phi(x,y)=0 \leadsto \) solve

1) \( \Phi_x = 3y+6x \)

2) \( \Phi_y = 6y+3x \)

\( \frac{d\Phi}{dx} = 3y+6x \leadsto \) integrate w.r.t \( x \)

\[ \begin{aligned} \Phi(x,y) &= \int (3y+6x)\,dx + C(y) \\ &= 3xy + 3x^2 + C(y) \end{aligned} \]

Plug in \(\phi(x,y) = 3xy + 3x^2 + C(y)\):

\[ \frac{\partial}{\partial y} \left[ 3xy + 3x^2 + C(y) \right] = 6y + 3x \] \[ \cancel{3x} + \frac{dC}{dy} = 6y + \cancel{3x} \] \[ \frac{dC}{dy} = 6y \implies C(y) = 3y^2 + K \]

\(\phi(x,y) = 3xy + 3x^2 + 3y^2 + K = 0\) is the general solution to

\[ \underbrace{(3y + 6x)}_{\phi_x} + \underbrace{(6y + 3x)}_{\phi_y} \frac{dy}{dx} = 0 \]

eg:

\[ \underbrace{(y^2 + \cos x)}_{N} \left(\frac{dx}{dy}\right) + \underbrace{(2xy + \sin y)}_{M} = 0 \] \[ \left. \begin{aligned} M_y &= 2x + \cos y \\ N_x &= -\sin x \end{aligned} \right\} \text{Not equal} \]

Observe! you have \( \frac{dx}{dy} \) not \( \frac{dy}{dx} \rightsquigarrow x, y \text{ are flipped} \)

Rewrite:

\[ \underbrace{(y^2 + \cos x)}_{M} + \underbrace{(2xy + \sin y)}_{N} \frac{dy}{dx} = 0 \] \[ M_y = 2y, \quad N_x = 2y \quad \rightsquigarrow \quad \text{Exact} \]

\[ (y^2 + \cos x) + (2xy + \sin y)\frac{dy}{dx} = 0 \]

1. \( \phi_x = y^2 + \cos x \)
2. \( \phi_y = 2xy + \sin y \)

From (1):

\[ \frac{\partial \phi}{\partial x} = y^2 + \cos x \implies \text{Integrate w.r.t } x \]

\[ \phi(x, y) = x y^2 + \sin x + C(y) \]

Using (2) \( \implies \)

\[ 2xy + \sin y = \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} \left[ x y^2 + \sin x + C(y) \right] \]

\[ \cancel{2xy} + \sin y = \cancel{2xy} + 0 + \frac{dC}{dy} \]

\[ \frac{dC}{dy} = \sin y \] \[ \Rightarrow C(y) = \int \sin y \, dy + k \] \[ = -\cos y + k \]

\( \Phi(x, y) = xy^2 + \sin x - \cos y + k = 0 \) is the general solution to \[ (y^2 + \cos x) + (2xy + \sin y)\frac{dy}{dx} = 0. \]