Separation of Variables

Given the differential equation
 

$M(x, y)\,dx + N(x, y)\,dy = 0$   ←   Equation (1)

 

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

$f(x)\,dx + f(y)\,dy = 0$   ←   Equation (2)

 

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
 

$\displaystyle \int f(x)\,dx + \int f(y)\,dy = c$

 

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
 

$M \, dx + N \, dy = 0$   ←   Equation (1)

 

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
 

$y = vx \,\, \text{and} \,\, dy = v \, dx + x\, dv$

or

$x = vy \,\, \text{and} \,\, dx = v \, dy + y\, dv$

 
The substitution above will lead to variables separable differential equation.
 

Exact Equations

The differential equation
 

$M(x, y) \, dx + N(x, y) \, dy = 0$

 

is an exact equation if
 

$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$

 

Steps in Solving an Exact Equation

  1. Let $\, \dfrac{\partial F}{\partial x} = M$.
  2. 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.

  3. Differentiate partially in terms of y the result in Step 2 holding x as constant.
  4. Equate the result in Step 3 to N and collect similar terms.
  5. Integrate the result in Step 4 with respect to y, holding x as constant.
  6. 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
 

$\dfrac{dy}{dx} + P(x) \, y = Q(x).$

 

The general solution of equation in this form is
 

$\displaystyle ye^{\int P\,dx} = \int Qe^{\int P\,dx}\,dx + C$

 

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,

$\displaystyle ye^{\int P\,dx} = \int Qe^{\int P\,dx}\,dx + C$