Integrating Factor Method

The integrating factor method is a standard technique for solving first-order linear ordinary differential equations of the form $$ y' + A(x)\, y = B(x) $$ It involves multiplying both sides of the equation by a specially chosen function known as the integrating factor: $$ M(x) = e^{\int A(x)\, dx} $$

The key idea is to turn the left-hand side of the equation into the derivative of a product, by applying the product or quotient rule in reverse:

$$ f'g + fg' = (f \cdot g)' $$

$$ \frac{f'g - fg'}{g^2} = \left( \frac{f}{g} \right)' $$

Once written as a derivative, the equation can be integrated directly with respect to \( x \) to find the solution.

    A Worked Example

    Consider the differential equation:

    $$ y' - \frac{2y}{x} = 0 $$

    This is a first-order homogeneous linear equation, which fits the standard form \( y' + A(x)\, y = B(x) \), with \( A(x) = -\frac{2}{x} \) and \( B(x) = 0 \).

    We begin by computing the integrating factor:

    $$ M(x) = e^{\int A(x)\, dx} $$

    $$ M(x) = e^{\int -\frac{2}{x}\, dx} $$

    $$ M(x) = e^{-2 \int \frac{1}{x}\, dx} $$

    Since \( \int \frac{1}{x}\, dx = \log x \), we obtain:

    $$ M(x) = e^{-2 \log x} = x^{-2} $$

    Equivalently:

    $$ M(x) = \frac{1}{x^2} $$

    Now we multiply both sides of the original equation by the integrating factor:

    $$ \left[ y' - \frac{2y}{x} \right] \cdot \frac{1}{x^2} = 0 $$

    Expanding the left-hand side:

    $$ \frac{y'}{x^2} - \frac{2y}{x^3} = 0 $$

    Bringing to common denominator:

    $$ \frac{y' x^3 - 2y x^2}{x^5} = 0 $$

    Factoring out \( x \):

    $$ \frac{x (y' x^2 - 2y x)}{x^5} = 0 $$

    Reducing:

    $$ \frac{y' x^2 - 2y x}{x^4} = 0 $$

    This expression is the result of applying the quotient rule in reverse, where \( f = y \) and \( g = x^2 \):

    $$ \left( \frac{y}{x^2} \right)' = 0 $$

    We now integrate both sides with respect to \( x \):

    $$ \int \left( \frac{y}{x^2} \right)' dx = \int 0 \, dx $$

    Which yields:

    $$ \frac{y}{x^2} = c $$

    Finally, solving for \( y \):

    $$ y = c \cdot x^2 $$

    This is the general solution of the original differential equation.

    And so on.

     

     
     

    Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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    Differential Equations

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