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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 $${\beta}$$ :

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

### First Order Clairaut

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

### Laplace Transform

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

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