`star` 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."}`

`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

Class Chapter 9 Example 2

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

Class Chapter 9 Example 2

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.