A function which satisfies the given differential equation is called its solution.

$\frac {dy}{dx} =x$

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

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

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.

(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

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

**Notes**- Differential equations
- General and Particular Solutions of Differential equations
- Method of solving for Differential equations

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