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1.6 Substitution Methods and Exact Eq. (Continued)

From calculus, if f(x, y) = c (level curve), where y is a function of x:

The total differential (or exact) of f is:

\[ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 0 \quad \text{or} \quad f_x dx + f_y dy = 0 \]

If f(x, y) is a well-behaving function, then:

\[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \quad \text{or} \quad f_{xy} = f_{yx} \]

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If we see a differential eq. in the form:

Exact Differential Eq.

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

And if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), then this matches the form:

\[ \left[ \frac{\partial f}{\partial x} \right] dx + \left[ \frac{\partial f}{\partial y} \right] dy = 0 \implies f(x, y) = c \]\[ \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \]

MUST have an implicit solution \( f(x, y) = c \) where:

\[ \frac{\partial f}{\partial x} = M \]\[ \frac{\partial f}{\partial y} = N \]

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Example: Exact Differential Equations

\[ (3x^2 + 2y^2)dx + (4xy + 6y^2)dy = 0 \]

Let \( M = 3x^2 + 2y^2 \) and \( N = 4xy + 6y^2 \).

Check if equation is exact: \( M_y = N_x \)?

\[ \begin{aligned} \frac{\partial}{\partial y} M &= \frac{\partial}{\partial y}(3x^2 + 2y^2) = 4y \\ \frac{\partial}{\partial x} N &= \frac{\partial}{\partial x}(4xy + 6y^2) = 4y \end{aligned} \]

These match, so the equation is exact.

Because it is exact, it has an implicit solution:

\[ f(x,y) = c \]

where:

\[ \begin{cases} \frac{\partial f}{\partial x} = M \\ \frac{\partial f}{\partial y} = N \end{cases} \]

Solve for \( f \).

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\[ \begin{aligned} \frac{\partial f}{\partial x} &= 3x^2 + 2y^2 \quad \text{--- (1)} \\ \frac{\partial f}{\partial y} &= 4xy + 6y^2 \quad \text{--- (2)} \end{aligned} \]

Pick one of them to integrate, here, let's pick (1).

From (1):

\[ f = \int (3x^2 + 2y^2) dx = x^3 + 2y^2x + g(y) \quad \text{--- (3)} \]

Note: \( y \) is a constant because this is reversing \( \frac{\partial}{\partial x} \).

Note: \( g(y) \) is constant in \( \frac{\partial}{\partial x} \).

We need to know \( g(y) \), so take the partial with respect to \( y \) of (3) and it should equal (2).

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From (3)

\[ \frac{\partial f}{\partial y} = 4yx + \frac{dg}{dy} = 4xy + 6y^2 \]

Note: \( 4xy + 6y^2 \) is \( N \) from (2)

So,

\[ \frac{dg}{dy} = 6y^2 \quad \implies \quad g(y) = 2y^3 + C_1 \]

Then,

\[ f(x, y) = x^3 + 2y^2x + 2y^3 + C_1 \]

Solution is implicitly given as \( f(x, y) = C \)

\[ x^3 + 2y^2x + 2y^3 = C \]

Absorbed the \( C_1 \) from \( f \)

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Exact Equations

Exact equations often come as \( M dx + N dy = 0 \), but they don't have to be expressed that way.

Example

\[ \frac{dy}{dx} = -\frac{y}{x} \]

This equation is:

separable
homogeneous
linear
exact

Rearranging the equation:

\[ x dy = -y dx \]\[ y dx + x dy = 0 \]

Here, \( M = y \) and \( N = x \).

Checking for exactness:

\[ \begin{cases} M_y = 1 \\ N_x = 1 \end{cases} \]

These match, so the equation is exact.

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Solving 2nd-Order Equations via Substitution

Some 2nd-order equations can be solved as 1st-order using substitutions.

First Type: No \(x\) in the Equation

\[ y'' = -y \]

Transform into a 1st-order with \(y\) as the independent variable.

Let \( p = \frac{dy}{dx} = y' \)

Then \[ y'' = \frac{dp}{dx} = \frac{dp}{dy} \frac{dy}{dx} = p \frac{dp}{dy} \]

Note: \(y\) is the independent variable here.

\[ p \frac{dp}{dy} = -y \]

This is first order in \(p\) and \(y\), and it is separable.

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\[ p \, dp = -y \, dy \] \[ \frac{1}{2} p^2 = -\frac{1}{2} y^2 + C \] \[ p^2 = -y^2 + C \] \[ p = \sqrt{-y^2 + C} = \frac{dy}{dx} \]

Rearranging for integration:

\[ \frac{dy}{dx} = \sqrt{-y^2 + C} \quad \text{(separable)} \]

[Note: A crossed-out integral expression follows in the original notes: \( y = \int p \, dx = \int \sqrt{-y^2 + C} \)]

Second Type: No \(y\) in the Equation

Keep \(x\) as the independent variable.

Let \( p = \frac{dy}{dx} = y' \)

Then \[ y'' = \frac{dp}{dx} \]

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Solving a Second-Order Differential Equation

Starting with the second-order differential equation:

\[ xy'' + y' = x \]

Let \( p = y' \). Substituting this into the equation gives a first-order differential equation in terms of \( p \) and \( x \):

\[ x \frac{dp}{dx} + p = x \]

1st order in \( p \) and \( x \)

Dividing by \( x \) to put the equation in standard linear form:

\[ \frac{dp}{dx} + \frac{1}{x} p = 1 \]

linear

Integrating Factor

Calculate the integrating factor \( I \):

\[ I = e^{\int \frac{1}{x} dx} = e^{\ln x} = x \]

Solving for \( p \)

Applying the integrating factor:

\[ \frac{d}{dx}(xp) = x \]

Integrating both sides with respect to \( x \):

\[ xp = \frac{1}{2}x^2 + c_1 \]

Solving for \( p \) (which is \( y' \)):

\[ p = \frac{1}{2}x + c_1 x^{-1} = y' \]

Final Solution for \( y \)

Integrating \( y' \) to find the general solution for \( y \):

\[ y = \frac{1}{4}x^2 + c_1 \ln x + c_2 \]