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General and Particular Solutions of Differential equations




General and Particular Solutions of Differential equations

Solutions of Differential Equations
A function which satisfies the given differential equation is called its solution.

Example
$\frac {dy}{dx} =x$
Now the solution of this differential equation would be any function y=f(x) that will satisfy it i.e., when the function is substituted for the unknown y (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S

Lets take the function as
$y=\frac {x^2}{2} + C$ where C is arbitrary constants
Substituting this function in differential equation , we find that
LHS=RHS

Lets put the value of constant C as 1 , then the function $y=\frac {x^2}{2} + 1$
Substituting this function in differential equation , we find that
LHS=RHS
So this is also a solution of the differential equation
So First equation was a general solutions and second one was a particular function

General Solution
The general solution of a differential equation involves arbitrary constants and represents a family of solutions i.e family of curves

Particular Solution
The particular solution is derived from the general solution by assigning specific values to the arbitrary constants, usually based on initial conditions or boundary values.

How to form Differential equations when general solutions is given

A Differential equation from a given function can be obatined by differentiating function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants

Example
(a)Family of Circles at origin
$x^2 + y^2 = a^2$
$2x + 2y \frac {dy}{dx}=0$
$x + y\frac {dy}{dx}=0$
This is the Differential equations

(b) Family of lines crossing through origin
$y=mx$
$\frac {dy}{dx} = m$
Substituting the value of m from original equation
$\frac {dy}{dx} = \frac {y}{x}$
or $x \frac {dy}{dx} -y=0$
This is the Differential equations

(c) Family of lines
$y=mx+c$
$\frac {dy}{dx} = m$
$\frac {d^2y}{dx^2} = 0

General Steps for Formation (a)If the given family F1 of curves depends on only one parameter then it is represented by an equation of the form
F1(x, y, a) = 0 ... (1)
Differentiating equation (1) with respect to x, we get an equation involving
y′, y, x, and a, i.e.,
g (x, y, y′, a) = 0 ... (2)
The required differential equation is then obtained by eliminating a from equations
(1) and (2) as
F(x, y, y′) = 0 ... (3)
(b) If the given family F2 of curves depends on the parameters a, b (say) then it is represented by an equation of the from
F2 (x, y, a, b) = 0 ... (4)
Differentiating equation (4) with respect to x, we get an equation involving
y′, x, y, a, b, i.e.,
g (x, y, y′, a, b) = 0 ... (5)
But it is not possible to eliminate two parameters a and b from the two equations and so, we need a third equation. This equation is obtained by differentiating
equation (5), with respect to x, to obtain a relation of the form h (x, y, y′, y″, a, b) = 0
The required differential equation is then obtained by eliminating a and b from equations (4), (5) and (6) as
F (x, y, y′, y″) = 0


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