PAGE 1

Exact Diff. Eqs. (Continued)

\( M(x,y) + N(x,y) y' = 0 \quad \text{or} \quad M(x,y)\,dx + N(x,y)\,dy = 0 \)

Such that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) or \( M_y = N_x \)

Exactness Criterion: By Clairaut's Theorem on the equality of mixed partial derivatives, a differential form is exact if and only if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) on a simply connected domain.

Solution: \( f(x,y) = C \) such that \( \frac{\partial f}{\partial x} = M \) and \( \frac{\partial f}{\partial y} = N \)

Given random 1st-order eq., it is likely not exact

but we have a way to make it exact

Integrating Factors: If an equation is not exact, we seek an integrating factor \( \mu(x,y) \) such that multiplying through yields an exact equation: \( \frac{\partial}{\partial y}[\mu M] = \frac{\partial}{\partial x}[\mu N] \).

for example,

\( \underbrace{e^x}_{M} + \underbrace{(e^x \cot y + 2y \csc y)}_{N} y' = 0 \)

\( \frac{\partial M}{\partial y} = 0 \)

\( \frac{\partial N}{\partial x} = e^x \cot y \)

clearly not exact

PAGE 2

Integrating Factors

if we multiply that equation by \(\sin y\)

\[\underbrace{e^x \sin y}_{M} + \underbrace{(e^x \cos y + 2y)}_{N} y' = 0\]

\[\frac{\partial M}{\partial y} = e^x \cos y \qquad \frac{\partial N}{\partial x} = e^x \cos y\]

it is now exact and we can (in principle) solve.

that \(\sin y\) is called an integrating factor (just like the one in 1st-order linear)

Note on Integrating Factors: In first-order linear differential equations, we multiply by an integrating factor \(I(x) = e^{\int P(x)\,dx}\) to produce an exact product rule. For non-exact equations of the form \(M(x,y) + N(x,y)y' = 0\), multiplying by \(\mu(x,y)\) aims to make the equation satisfy \(\frac{\partial(\mu M)}{\partial y} = \frac{\partial(\mu N)}{\partial x}\).

General Existence of an Integrating Factor

Can we always find one? If so, what is it?

Given \[M(x, y) + N(x, y)y' = 0 \quad \text{but} \quad M_y \neq N_x\]

want to multiply by

\[\underbrace{\mu(x, y)}_{\substack{\text{can depend} \\ \text{on BOTH } x \text{ and } y}}\]

such that the resulting eq. is exact

PAGE 3

Finding an Integrating Factor: Exactness Condition

Multiplying the differential equation by an integrating factor \(\mu(x,y)\):

\[ \underbrace{[\mu(x,y) M(x,y)]}_{\text{new } M} + \underbrace{[\mu(x,y) N(x,y)]}_{\text{new } N} y' = 0 \]

Note on Exactness: For a differential equation in the form \(M + N y' = 0\) to be exact, we require \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). When multiplied by an integrating factor \(\mu(x,y)\), this condition must hold for the modified functions \(\mu M\) and \(\mu N\).

\[ \text{want: } \frac{\partial}{\partial y} [\mu(x,y) M(x,y)] = \frac{\partial}{\partial x} [\mu(x,y) N(x,y)] \]

Expanding via the product rule:

\[ \mu M_y + \mu_y M = \mu N_x + \mu_x N \]

\[ \mu_y M - \mu_x N + \mu(M_y - N_x) = 0 \]

solve this for \(\mu(x,y)\)

this is a partial diff. eq. (PDE)

very hard to solve for general \(\mu(x,y)\)

ANY \(\mu\) that satisfies the eq. can be used as an integrating factor

Simplifying assumption:

assume \(\mu = \mu(x)\) function of \(x\) only
or \(\mu = \mu(y)\) function of \(y\) only

Simplification Strategy: Assuming \(\mu\) is a function of only \(x\) implies \(\mu_y = 0\), reducing the partial differential equation (PDE) to a first-order linear ordinary differential equation (ODE) in \(x\), which is straightforward to solve.

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Integrating Factor as a Function of \( y \): \( \mu = \mu(y) \)

If we go with \( \mu = \mu(y) \), then \( \mu_x = 0 \) and:

\[ \mu_y = \frac{\partial \mu}{\partial y} = \frac{d\mu}{dy} \] PDE \( \to \) ODE

When \( \mu \) depends solely on the single variable \( y \), the partial derivative \( \frac{\partial \mu}{\partial y} \) simplifies to an ordinary derivative \( \frac{d\mu}{dy} \), reducing the partial differential equation (PDE) to an ordinary differential equation (ODE).

Equation on last page becomes:

\[ M \frac{d\mu}{dy} + \mu (M_y - N_x) = 0 \]

Rewrite:

\[ \frac{d\mu}{dy} = -\frac{(M_y - N_x)}{M} \mu \] Separable!

This ODE can be solved by separation of variables provided that the expression \( \frac{M_y - N_x}{M} \) is a function of \( y \) only (or a constant).

Example

\[ (e^{2x} + y - 1) \, dx - (1) \, dy = 0 \]

Testing for exactness with \( M = e^{2x} + y - 1 \) and \( N = -1 \):

\[ \frac{\partial}{\partial y}(e^{2x} + y - 1) \neq \frac{\partial}{\partial x}(-1) \] NOT exact.

Evaluating the partial derivatives gives \( M_y = 1 \) and \( N_x = 0 \). Since \( M_y \neq N_x \), the differential equation is not exact.

Multiply by \( \mu \) to make it exact:

\[ \mu (e^{2x} + y - 1) \, dx - \mu \, dy = 0 \]

We want:

\[ \frac{\partial}{\partial y}\bigl[\mu (e^{2x} + y - 1)\bigr] = \frac{\partial}{\partial x}[-\mu] \]

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Integrating Factor: Solving for \(\mu(x)\)

assume \(\mu = \mu(x)\)

left side : \(\mu\) is a “constant”

\[ \mu = -\frac{d\mu}{dx} \]

solve for \(\mu\) (separable, linear)

Step-by-step separation of variables: Rewrite as \(\frac{1}{\mu}\,d\mu = -dx\). Integrating both sides gives \(\ln|\mu| = -x + C_1\), so \(\mu(x) = C e^{-x}\).

\[ \mu = C e^{-x} \]

choose ANY \(C \neq 0\)

\[ \mu = e^{-x} \]

Setting the constant \(C = 1\) provides the simplest non-zero integrating factor.

Transformation to an Exact Equation

original eq. :

\[ (e^{2x} + y - 1)\,dx - dy = 0 \]

multiply by \(\mu = e^{-x}\)

\[ (e^x + y e^{-x} - e^{-x})\,dx - e^{-x}\,dy = 0 \]\[ y e^{-x}\,dx - e^{-x}\,dy = 0 \]

\[ \begin{aligned} \frac{\partial}{\partial y}(y e^{-x}) &= e^{-x} \\[6pt] \frac{\partial}{\partial x}(-e^{-x}) &= e^{-x} \end{aligned} \]

exact !

Verification of Exactness: For the differential equation \(M(x,y)\,dx + N(x,y)\,dy = 0\), exactness requires \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). Here, \(M = e^x + y e^{-x} - e^{-x}\) and \(N = -e^{-x}\). Differentiating yields \(\frac{\partial M}{\partial y} = e^{-x}\) and \(\frac{\partial N}{\partial x} = e^{-x}\). Because they are equal, the multiplied equation is exact.

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Solution: Potential Function Method

Solution: \( f(x, y) = C \) such that

\[ \begin{aligned} \frac{\partial f}{\partial x} &= y e^{-x} && (\text{"new" } M) \\[8pt] \frac{\partial f}{\partial y} &= -e^{-x} && (\text{"new" } N) \end{aligned} \]

\[ \begin{aligned} \frac{\partial f}{\partial x} &= y e^{-x} && \text{--- ①} \\[8pt] \frac{\partial f}{\partial y} &= -e^{-x} && \text{--- ②} \end{aligned} \]

integrate ①

\[ f = \int y e^{-x} \, dx = -y e^{-x} + h(y) \]

Integrating \( \frac{\partial f}{\partial x} \) with respect to \( x \) treats \( y \) as a constant. The constant of integration may depend on \( y \), represented by the arbitrary function \( h(y) \).

\[ \frac{\partial f}{\partial y} = -e^{-x} + \frac{dh}{dy} = \underbrace{-e^{-x}}_{\text{②}} \]

so, \( h = \text{constant} \)

Differentiating \( f(x,y) \) with respect to \( y \) yields \( -e^{-x} + h'(y) \). Comparing this to equation ② gives \( h'(y) = 0 \), which implies \( h(y) \) is a constant.

Solution:

\[ -y e^{-x} = C \]

or \( y = C e^x \)

Alternative Method Observation

orig. eq.

\[ e^{2x} + y - 1 - y' = 0 \longrightarrow y' - y = e^{2x} - 1 \quad \text{linear!} \]

think about the most efficient method!

Rearranging the original differential equation into standard linear form \( y' + P(x)y = Q(x) \) demonstrates that it could have been solved directly using a standard linear integrating factor \( I(x) = e^{\int -1 dx} = e^{-x} \), which is often much quicker than solving as an exact differential equation.

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Are linear eqs. exact ?

\[ y' + p(x)y = g(x) \]

“multiply” by \( dx \)

\[ dy + p(x)y \, dx = g(x) \, dx \]

\[ \underbrace{\bigl[ p(x)y - g(x) \bigr]}_{M} \, dx + \underbrace{1}_{N} \, dy = 0 \]

Rewriting the first-order linear equation into the standard differential form \( M(x,y)\,dx + N(x,y)\,dy = 0 \) identifies \( M(x,y) = p(x)y - g(x) \) and \( N(x,y) = 1 \).

\[ \text{exact : } \frac{\partial}{\partial y} \bigl[ p(x)y - g(x) \bigr] = \frac{\partial}{\partial x} [1] = 0 \]

exact if \( p(x) = 0 \)

else, not

Since \( \frac{\partial M}{\partial y} = p(x) \) and \( \frac{\partial N}{\partial x} = 0 \), the equality \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) holds if and only if \( p(x) = 0 \). Thus, general linear equations are not exact unless the coefficient of \( y \) is zero.

how is the integrating factor for linear related to the one to make eq. exact ?

\[ \text{linear: } \mu = e^{\int p(x)\,dx} = \mu(x) \]

Multiplying the non-exact differential equation \( [p(x)y - g(x)]\,dx + 1\,dy = 0 \) by \( \mu(x) = e^{\int p(x)\,dx} \) yields an exact equation where \( \frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N) \).

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Integrating Factor and Exact Differential Equations

multiply \( [ p(x)y - g(x) ]\,dx + dy = 0 \) by that

Context: A first-order linear differential equation \( \frac{dy}{dx} + p(x)y = g(x) \) can be written in differential form as \( [p(x)y - g(x)]\,dx + dy = 0 \). Multiplying through by the integrating factor \( \mu(x) = e^{\int p(x)\,dx} \) transforms it into an exact differential equation.

\[ (\mu p y - \mu g)\,dx + \mu\,dy = 0 \]

\[ \frac{\partial}{\partial y} (\mu p y - \mu g) = \mu p \]

\[ \begin{aligned} \frac{\partial}{\partial x} (\mu) &= \frac{d}{dx} \left( e^{\int p(x)\,dx} \right) = e^{\int p(x)\,dx} \frac{d}{dx} \left( \int p(x)\,dx \right) \\ &= \mu p \end{aligned} \]

Exactness Criterion: For an equation \( M(x, y)\,dx + N(x, y)\,dy = 0 \) to be exact, we require \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Here, \( M = \mu p y - \mu g \) and \( N = \mu \). Differentiating gives \( \frac{\partial M}{\partial y} = \mu p \) and \( \frac{\partial N}{\partial x} = \mu p \). Because these partial derivatives are equal, the equation is exact.

So, the integrating factor for linear eq. is the same that makes an eq. exact!