# General and Particular Solution of a Differential Equation

## General and Particular Solution of a Differential Equation

We have solved the equation of the type:

x² + 1 = 0 … (1)

sin²x – cosx = 0 … (2)

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

now consider the differential equation d²y/ dx² + y = 0 … (3)

the solution of this differential equation is a function φ that will satisfy it i.e., when the function φ is substituted for the unknow y 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

y = φ (x) = a sin (x + b) … (4)

where a, b ϵ 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 = π/4, then we get a function

y = φ₁ (x) = 2 sin (x + π/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).

Function φ consists of two arbitrary constant a, b and it is called general solution of the given differential equation. Whereas function φ₁ contains no arbitrary constants but only the particular values of the parameters a and b hence is called a Particular solution of the given differential equation.

Example: verify that the function y = e⁻³x is a solution of the differential equation $$\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}-6y=0$$.

Solution: Given that y = e⁻³x

Differentiate with respect x

$$\frac{dy}{dx}=-3\times {{e}^{-3x}}$$ … (1)

Again, Differentiate with respect x

$$\frac{{{d}^{2}}y}{d{{x}^{2}}}=-3\times -3\times {{e}^{-3x}}$$,

$$\frac{{{d}^{2}}y}{d{{x}^{2}}}=9{{e}^{-3x}}$$ … (2)

From equation (1) and (2)

$$\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}-6y=0$$ (given that)

9e⁻³x – 3e⁻³x – 6y = 0

9e⁻³x – 3e⁻³x – 6e⁻³x = 0

9e⁻³x – 9e⁻³x = 0

L.H.S = R.H.S

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