## Separation of Variables

Given the differential equation

where *M* and *N* may be functions of both *x* and *y*. If the above equation can be transformed into the form

where *f*(*x*) is a function of *x* alone and *f*(*y*) is a function of *y* alone, equation (1) is called **variables separable**.

To find the general solution of equation (1), simply equate the integral of equation (2) to a constant *c*. Thus, the general solution is

## Homogeneous Functions

If the function *f*(*x*, *y*) remains unchanged after replacing *x* by *kx* and *y* by *ky*, where *k* is a constant term, then *f*(*x*, *y*) is called a ** homogeneous function**. A differential equation

is homogeneous in *x* and *y* if *M* and *N* are homogeneous functions of the same degree in *x* and *y*.

To solve for Equation (1) let

or

The substitution above will lead to variables separable differential equation.

## Exact Equations

The differential equation

is an exact equation if

**Steps in Solving an Exact Equation**

- Let $\, \dfrac{\partial F}{\partial x} = M$.
- Write the equation in Step 1 into the form

$\displaystyle \int \partial F = \int M \, \partial x$and integrate it partially in terms of

*x*holding*y*as constant. - Differentiate partially in terms of
*y*the result in Step 2 holding*x*as constant. - Equate the result in Step 3 to
*N*and collect similar terms. - Integrate the result in Step 4 with respect to
*y*, holding*x*as constant. - Substitute the result in Step 5 to the result in Step 2 and equate the result to a constant
*c*.

## Linear Equations of Order One

Linear equation of order one is in the form

The general solution of equation in this form is

**Derivation**

$\dfrac{dy}{dx} + Py = Q$

Use $\,e^{\int P\,dx}\,$ as integrating factor.

$e^{\int P\,dx} \dfrac{dy}{dx} + Pe^{\int P\,dx} \, y = Qe^{\int P\,dx}$

Multiply both sides of the equation by *dx*

$e^{\int P\,dx} \,dy + Pe^{\int P\,dx} \, y \, dx = Qe^{\int P\,dx}\, dx$

Let

$u = \int P\,dx$

$du = P\,dx$

Thus,

$e^u \,dy + ye^u \, du = Qe^u\, dx$

$d(e^u y) = Qe^u\, dx$

$\displaystyle d(e^u y) = \int Q e^u\, dx$

But $\,u = \int P\,dx\,$. Thus,