Mathematics General and Particular Solutions of a Differential Equation For CBSE-NCERT 2
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star General and Particular Solutions of a Differential Equation

### General and Particular Solutions of a Differential Equation

Cosider the following equation

color{orange} {x^2 +1 = 0}.............................(1)

color{orange} {sin^2x - cos x = 0} ...............(2)

=> Solution of equations (1) and (2) are numbers, real or complex, that will satisfy the given equation i.e., when that number is substituted for the unknown x in the given equation, L.H.S. becomes equal to the R.H.S..

=> Now consider the differential equation color{orange} {(d^2y)/(dy^2) + y = 0} ..........(3)

=> In contrast to the first two equations, the solution of this differential equation is a function φ that will satisfy it i.e., when the function φ is substituted for the unknown y (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S..

The curve y = φ (x) is called the solution curve (integral curve) of the given differential equation. Consider the function given by

color{orange} {y = φ (x) = a sin (x + b)}, ....................... (4)

where a, b ∈ R. When this function and its derivative are substituted in equation (3), L.H.S. = R.H.S.. So it is a solution of the differential equation (3).

=> Let a and b be given some particular values say a = 2 and b = pi/4, then we get a function

color{orange} {y = y = φ_1(x) = 2sin (x + pi/4)} .............................(5)

When this function and its derivative are substituted in equation (3) again L.H.S. = R.H.S.. Therefore φ1 is also a solution of equation (3).

=>The function φ consists of two arbitrary constants (parameters) a, b and it is called general solution of the given differential equation.

=> Whereas function φ_1 contains no arbitrary constants but only the particular values of the parameters a and b and hence is called a particular solution of the given differential equation.

color{red}{=>"The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation."}

color{red}{=>" The solution free from arbitrary constants i.e., the solution obtained from the general solution"}
color{red}{" by giving particular values to the arbitrary constants is called a particular solution of the differential equation."}
Q 3186180977

Verify that the function y = e^(- 3x) is a solution of the differential equation (d^y)/(dx^2) + (dy)/(dx) - 6y =0

Class Chapter 9 Example 2
Solution:

Given function is y = e^(-3x). Differentiating both sides of equation with respect to x , we get

(dy)/(dx) = -3e^(-3x) ...........................(1)

Now, differentiating (1) with respect to x, we have

(d^y)/(dx^2) = 9e^(-3x)

Substituting the values of (d^2y)/(dx^2), (dy)/(dx) and y in the given differential equation, we get

L.H.S = 9 e^(-3x) + (–3e^-(3x)) – 6.e^(- 3x) = 9 e^(-3x) – 9 e^(- 3x) = 0 = R.H.S..

Therefore, the given function is a solution of the given differential equation.
Q 3116180979

Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution of the differential equation (d^2y)/(dx^2) + y = 0

Class Chapter 9 Example 2
Solution:

The given function is

y = a cos x + b sin x ............... (1)

Differentiating both sides of equation (1) with respect to x, successively, we get

(dy)/(dx) = - a sin x + b cos x

(d^2y)/(dx^2) = - a cos x - b sinx

Substituting the values of (d^2y)/(dx^2) and y in the given differential equation, we get

L.H.S. = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R.H.S..

Therefore, the given function is a solution of the given differential equation.