Second Order Differential Equation

Second Order Homogeneous Differential Equation
The general form of Second order differential equation is given by, a d^y/dx^2 + b dy/dx + cy= f(x) where a, b, c are some constants and f(x) is a function of x.  The equation is homogenous when f(x) equals zero, ay’’ + by’ + cy = 0 is a second order homogenous differential equation. For instance, d^2y/dx^2 -2 dy/dx + 3y = 0 is an example for a second order homogenous differential equation.

Second Order Non Linear differential Equation
The general form of second order nonlinear differential equation is given by,  y'' + p(t)y'' + q(t) y = f(x). The general solution is given by y = y(p) x + y(h)x where the particular solution of non-homogenous is y(p) x and the general solution of the associated homogenous equation is y(h)x.  Second order nonlinear differential equation can also be written as (a) d^2y/dx^2 + (b) dy/dx + (c)y = f(x) where a, b, c are constants and f(x) is the function of x.

In general, a y’’ + b y’ + c y = 0 is the homogeneous second order differential equation where the constants are a, b and c and also a does not equal zero. Using this equation we can arrive to an auxiliary equation or the characteristic equation that has roots which helps in solving second order differential equations. Let us consider the function y=e^(rx) an exponential function with r as constant. We know that the derivative of an exponential function is a product of itself and the constant, using this property we get, y’= r.e^(rx) and y’’= r^2.e^(rx). Let us substitute these expressions in the general form, that gives us, a[r^2.e^(rx)]+ b.[r. e^(rx)] + c [e^(rx)]= 0. Here e^(rx) never equals zero and hence y=e^(rx) is the solution of the equation. The auxillary equation is ar^2 +b r + c = 0 which is an algebraic equation for which the roots r1 and r2 can be found either by factoring or by using the quadratic formula. If the roots are unequal and real then the general solution is given by, y = c1e^(r1x) + c2e^(r2x)

Second order differential equation examples
Solve the given equation 3y’’ +y’- y = 0
Here the auxiliary equation would be,  3r^2 + r -1 =0.
Using the quadratic formula, [-b (+/-) sqrt(b^2-4ac)]/2a ; a = 3, b = 1, c = -1
We get, {-1(+/-)sqrt[1^2 – (4.3.-1)]}/2.1 = [-1(+/-)sqrt(13)]/6

The roots being real and distinct, the general solution is,
 y = c1e^[x/6(-1+sqrt(13)] + c2 e^[x/6(1+sqrt(13)]