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1.6 Substitution Methods and Exact Equation (part 1)

Many equations are separable, linear, or both.

Many more are neither, for example \[ \frac{dy}{dx} = (x+y)^2 \]

Some can be turned into separable, linear, or both by using substitutions.

Example

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

Let \( v = x + y \)

Transform eq. into one in \( v \) and \( x \) (no \( y \)).

From \( v = x + y \)

We get:

\[ \frac{dv}{dx} = \frac{d}{dx}x + \frac{d}{dx}y = 1 + \frac{dy}{dx} \]
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\[ \frac{dy}{dx} = (x+y)^2 \]

Substituting the expressions for \( \frac{dy}{dx} \) and \( (x+y) \):

\[ \frac{dv}{dx} - 1 = v^2 \]
\[ \frac{dv}{dx} = v^2 + 1 \quad \text{separable in } v \text{ and } x \]
\[ \frac{1}{v^2 + 1} dv = dx \]

Integrate:

\[ \int \frac{1}{v^2 + 1} dv = \int dx \]
\[ \tan^{-1}(v) = x + C \]
\[ v = \tan(x + C) = x + y \]
\[ y = \tan(x + C) - x \]
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Substitution Methods for Differential Equations

Any equation in the form

\[ \frac{dy}{dx} = f(ax + by + c) \quad a, b, c \text{ constants} \]

can be solved by using the substitution \( V = ax + by + c \).

Example

For example,

\[ \frac{dy}{dx} = \sqrt{3x - 2y + 1} = f(3x - 2y + 1) \]

Sub: \( V = 3x - 2y + 1 \)

Then proceed as in the previous example.

Homogeneous Equations

A first-order eq. is said to be homogeneous if it can be written as

\[ \frac{dy}{dx} = f\left(\frac{y}{x}\right) \]

Note: "homogeneous" has different meanings in different contexts.

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Example: Homogeneous Substitution

For example,

\[ \frac{dy}{dx} = \frac{2x - y}{x + 7y} \]
  • not separable
  • not linear

If we divide top and bottom on right side by \( x \):

\[ \frac{dy}{dx} = \frac{2 - (\frac{y}{x})}{1 + 7(\frac{y}{x})} = f\left(\frac{y}{x}\right) \]

So eq. is homogeneous.

homogeneous: use sub \( v = \frac{y}{x} \)

Get \( v \) into eq., get \( y \) out.

\[ \frac{dy}{dx} = \frac{2 - (\frac{y}{x})}{1 + 7(\frac{y}{x})} \]

\( v = \frac{y}{x} \)

\( y = vx \)

Use this to find \( \frac{dy}{dx} \) in terms of \( v \) and \( \frac{dv}{dx} \).

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Substitution Method for Differential Equations

Let us define a substitution where:

\[ y = vx \]

Note: Here, \( v \) is a function of \( x \).

To find the derivative, we must use the product rule:

\[ \frac{dy}{dx} = v \cdot 1 + x \cdot \frac{dv}{dx} = v + x \frac{dv}{dx} \]

Applying the Substitution

The original equation (e.g.) becomes:

\[ v + x \frac{dv}{dx} = \frac{2 - v}{1 + 7v} \]

Isolating the term with the derivative:

\[ x \frac{dv}{dx} = \frac{2 - v}{1 + 7v} - v \]\[ x \frac{dv}{dx} = \frac{2 - v}{1 + 7v} - \frac{v(1 + 7v)}{1 + 7v} \]\[ x \frac{dv}{dx} = \frac{2 - v - v - 7v^2}{1 + 7v} \]

Simplifying the numerator:

\[ x \frac{dv}{dx} = \frac{2 - 2v - 7v^2}{1 + 7v} \]

This equation is now separable in \( v \) and \( x \).

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Solving the Separable Equation

Separating the variables and integrating both sides:

\[ \int \frac{1 + 7v}{2 - 2v - 7v^2} \, dv = \int \frac{1}{x} \, dx \]

Integration by Substitution

Let us use a substitution for the left-hand side integral:

\[ u = 2 - 2v - 7v^2 \]\[ du = (-2 - 14v) \, dv \]\[ du = -2(1 + 7v) \, dv \]

Substituting these into the integral:

\[ -\frac{1}{2} \int \frac{1}{u} \, du = \int \frac{1}{x} \, dx \]

Final Solution

Integrating both sides yields:

\[ -\frac{1}{2} \ln u = \ln x + C \]\[ \ln u = -2 \ln x + C \]\[ u = e^{-2 \ln x + C} = e^{\ln x^{-2}} \cdot e^C = Cx^{-2} \]

Substituting back for \( u \):

\[ 2 - 2v - 7v^2 = Cx^{-2} \]
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\[ 2 - 2 \left( \frac{y}{x} \right) - 7 \left( \frac{y}{x} \right)^2 = cx^{-2} \]

make it prettier, multiply by \( x^2 \)

\( 2x^2 - 2xy - 7y^2 = c \)

Bernoulli Differential Eq.

\[ y' + P(x)y = Q(x)y^n \]

\( n \neq 0, n \neq 1 \)

eq. becomes linear

eq. becomes linear and/or separable

if \( n = 1 \), \( y' + Py = Qy \)

\( y' + (P - Q)y = 0 \) linear

where \( R = P - Q \)

\( y' = -Ry \) might be separable

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for Bernoulli, use sub \( v = y^{1-n} \)

example

\[ x^2 y' + xy = y^2 \]

make coefficient of \( y' \) 1

\[ y' + \frac{1}{x}y = \frac{1}{x^2}y^2 \implies n = 2 \]

let \( v = y^{1-n} = y^{-1} \)

rewrite eq. in terms of \( v \) and \( v' \), eliminating \( y \) and \( y' \)

from \( v = y^{-1} \)

\[ v' = -y^{-2}y' \quad \text{so} \quad y' = -y^2 v' \]
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Solving a Bernoulli Differential Equation

Returning to the original equation:

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

Divide by \(-y^2\):

\[ v' - \frac{1}{x} \cdot \left( \frac{1}{y} \right) = -\frac{1}{x^2} \]
\[ y^{-1} = v \]
\[ v' - \frac{1}{x}v = -\frac{1}{x^2} \]

This is linear in \(v\) and \(x\).

Finding the Integrating Factor

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

Multiplying the linear equation by the integrating factor:

\[ x^{-1} v' - \frac{1}{x^2} v = -\frac{1}{x^3} \]
\[ \frac{d}{dx}(x^{-1}v) = -\frac{1}{x^3} \]
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Integration and Back-Substitution

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

\[ x^{-1}v = \frac{1}{2x^2} + C \]

Solving for \(v\):

\[ v = \frac{1}{2x} + Cx \]
\[ v = y^{-1} = \frac{1}{y} \]

Substituting back for \(y\):

\[ \frac{1}{y} = \frac{1}{2x} + Cx \]
\[ y = \frac{1}{\frac{1}{2x} + Cx} = \frac{2x}{1 + 2Cx^2} = \frac{2x}{1 + Cx^2} \]
\[ y = \frac{2x}{1 + Cx^2} \]