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Differential Equations Cheat Sheet

First Order Separable

$${\frac{dy}{dx} = f(x) \ g(y)}$$

Manipulate the expression into the form:

$${g(y) \ dy = f(x) \ dx}$$

First Order Substitution

\({u = \frac{y}{x}}\) is called first order homogeneous. And substitute \({y' = u + x \ u'}\).

\({u = ax + by}\) is another common substitution. And substitute \({u' = a + b \ y'}\).

First Order Linear

The problem is linear because it is linear in y and the left side of the equation can be expressed as the derivative of y and an integrating factor.

$${\frac{dy}{dx} + f(x) \ y = g(x)}$$

First Order Homogeneous

Make a homogeneous substitution of the form \({v = \frac{y}{x}}\) and the resulting equation will be separable.

First Order Exact

$${Mx \ dx + Ny \ dy = 0}$$

$${M_{xy} = N_{yx}}$$

First Order Integrating Factor

Find an integrating factor \({u(x)}\) or \({u(y)}\) and multiply the factor to both sides such that the problem becomes exact.

For a function of the type \({u(x,y)}\) try \({x^m y^n}\) .

First Order Bernoulli

Use a Bernoulli substitution in the form \({v = y^{1-n}}\) where n is the power of y on the right side

of a First Order equation that is otherwise in linear form. With the substitution the problem becomes linear.

For a problem initially in the form:

$${\frac{dy}{dx} + f(x) \ y = g(x) \ y^n}$$

After the substitution the linear equation in v will be of the form:

$${\frac{dv}{dx} + (1 - n) \ f(x) \ v = (1 - n) \ g(x)}$$

2nd Order Homogeneous

$${ay'' + by' + cy = 0}$$

For the roots \({\alpha}\) and \({\beta}\) :

$${y = c_1 e^{\alpha t} + c_2 e^{\beta t}}$$

Higher Order Homogeneous

First Order Undetermined Coefficients

2nd Order Undetermined Coefficients

Higher Order Undetermined Coefficients

2nd Order Variation of Parameters

First Order Clairaut

$${y = xy' + f(y')}$$

Laplace Transform

$${\mathcal{L}\{f'(t)\} = sF(s) - f(0)}$$

$${\mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \dots - s f^{(n-1)}(0) - f^{(n)}(0)}$$