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2.6 (Continued)

exact: \( M(x,y) + N(x,y)y' = 0 \) such that \( M_y = N_x \)

solution is \( \psi(x,y) = c \) where \( \psi_x = M \) and \( \psi_y = N \)

Sometimes, we can make an equation exact by multiplying by an integrating factor like with linear eqs. (same idea but different process)

\[ e^x + (e^x \cot y + 2y \csc y)y' = 0 \]

\( M = e^x \) and \( N = e^x \cot y + 2y \csc y \)

\( M_y = 0 \)

\( N_x = e^x \cot y \)

\( M_y \neq N_x \) so not exact

but notice if we multiply both sides by \( \sin y \)

\[ e^x \sin y + (e^x \cos y + 2y)y' = 0 \]

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new \( M = e^x \sin y \)

new \( N = e^x \cos y + 2y \)

\( M_y = e^x \cos y \)

\( N_x = e^x \cos y \)

exact!

now find \( \psi \) such that \( \psi_x = e^x \sin y \), \( \psi_y = e^x \cos y + 2y \)

How to find \( \mu \) to make eq. exact?

\( M(x,y) + N(x,y)y' = 0 \) but \( M_y \neq N_x \)

multiply by \( \mu(x,y) \)

\[ \mu(x,y)M(x,y) + \mu(x,y)N(x,y)y' = 0 \]

to be exact

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

\[ \mu(x,y)M_y(x,y) + \mu_y(x,y)M(x,y) = \mu(x,y)N_x(x,y) + \mu_x(x,y)N(x,y) \]

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

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

Solve this partial diff. eq. (PDE) for \( \mu \)

not easy: much harder than the original (non-exact) ODE.
ANY \( \mu \) that satisfies this eq. can be used to make the eq. exact.

assume \( \mu \) is \( \mu(x) \) or \( \mu(y) \) only

then one of \( \mu_x \) or \( \mu_y \) goes away.

for example, if we say \( \mu = \mu(y) \)

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

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

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example

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

\( M = e^{2x} + y - 1 \)

\( N = -1 \)

\( M_y = 1 \)

\( N_x = 0 \)

not exact

find a \( \mu \) to make it exact

\[ \mu (e^{2x} + y - 1) - \mu y' = 0 \]is exact

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

make a choice: \( \mu = \mu(x) \) or \( \mu = \mu(y) \)

let's go with \( \mu = \mu(x) \)

on the left, \( \mu \) is "constant"

\[ \text{so, } \mu = - \frac{d\mu}{dx} \]separable

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Solving Differential Equations with Integrating Factors

\[ \frac{1}{\mu} d\mu = -dx \]

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

Since \( \mu \) is multiplied to both sides of the original equation, \( C \) is arbitrary. Choose any \( C \neq 0 \).

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

Applying the Integrating Factor

Back to \( (e^{2x} + y - 1) - y' = 0 \)

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

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

\[ (e^x + e^{-x}y - e^{-x}) + (-e^{-x})y' = 0 \]

Verification of Exactness

new \( M = e^x + e^{-x}y - e^{-x} \)
\( N = -e^{-x} \)

\( M_y = e^{-x} \)
\( N_x = e^{-x} \)

Finding the Potential Function \( \Psi \)

Find \( \Psi \):

\[ \Psi_x = e^x + e^{-x}y - e^{-x} \]

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

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This time, let's integrate \( \Psi_y = -e^{-x} \)

\[ \Psi = \int -e^{-x} dy = -e^{-x}y + h(x) \]

Determining \( h(x) \)

\[ \Psi_x = e^{-x}y + h'(x) \]

Must match \( M = e^x + ye^{-x} - e^{-x} \)

\[ h'(x) = e^x - e^{-x} \]

\[ h(x) = e^x + e^{-x} \]

No need for \( C \) here.

Final Solution

So, \( \Psi = -e^{-x}y + e^x + e^{-x} \)

Solution: \( \Psi = C \)

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

\[ e^{2x} - 1 - y = Ce^x \]

\[ y = e^{2x} - 1 + Ce^x \]

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let's look at the original eq. again

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

\[y' - y = e^{2x} - 1\]

hold up! it's linear!

\[\mu = e^{\int -1 dx} = e^{-x}\]

\[e^{-x}(y' - y) = e^{-x}(e^{2x} - 1)\]

\[\frac{d}{dx}(e^{-x}y) = e^x - e^{-x}\]

\[e^{-x}y = e^x + e^{-x} + c\]

\[y = e^{2x} + 1 + ce^x\]

same!