How to Find Particular Integral: Method, Formula, Examples

The particular integral of a differential equation is needed to find to solve the equation. In this article, lets learn how to find particular integral with formula and method.

Method of Finding Particular Integral (PI)

Let us consider a second order linear differential equation with constant coefficients:

$\dfrac{d^2y}{dx^2}+a\dfrac{dy}{dx}+by=Q(x)$ …(1)

where a and b constants and Q(x) is a function of x.

Letting $D \equiv \dfrac{d^n}{dx^n}$, we can rewrite the above equation (1) as follows.

D2y + aDy + by = Q(x) ⇒ (D2+aD+b)y = Q(x).

By definition, the particular integral (PI) of Equation (1) is given by

$\boxed{\text{PI} = \dfrac{1}{D^2+aD+b} Q(x)}$

Writing f(D) = D²+aD+b, we have:

$\boxed{\text{PI} = \dfrac{1}{f(D)} Q(x)}$

The method of finding particular integral of a differential equation is listed below:

Formula 1:

Particular Integral of e^ax

The particular integral formula of eax is given as follows: 1. $\dfrac{1}{f(D)} e^{ax} = \dfrac{e^{ax}}{f(a)}$ if f(a) ≠ 0. 2. $\dfrac{1}{(D-a)^r \phi(D)} e^{ax} = \dfrac{x^r}{r!} \dfrac{e^{ax}}{\phi(a)}$ if φ(a) ≠ 0.

Let us now learn how to apply these formulae.

Question 1: Solve $\dfrac{d^2y}{dx^2}-y=e^{2x}$.

Answer:

The equation can be written as (D²-1)y = e^2x whose complete solution is given by y = Complementary Function (CF) + Particular Integral (PI). That is,

y = CF + PI

$\boxed{\text{CF :}}$ The auxiliary equation is given by m²-1 =0

⇒ m² = 1.

⇒ m = 1, -1 (real and unequal roots)

∴ CF = Ae^x+Be^-x where A and B are arbitrary constants.

$\boxed{\text{PI :}}$ The particular integral is given by

$\dfrac{1}{D^2-1}e^{2x}$

= $\dfrac{e^{2x}}{2^2-1}$

= $\dfrac{e^{2x}}{3}$

Therefore, the complete solution of (D²-1)y = e^2x is given by y = CF + PI = Ae^x+Be^-x + $\dfrac{e^{2x}}{3}$ where A and B are arbitrary constants.

Question 2: Solve $\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}+y=e^{-x}$.

Answer:

The equation can be written as (D²+2D+1)y = e^-x whose complete solution is given by y = Complementary Function (CF) + Particular Integral (PI). That is,

y = CF + PI

$\boxed{\text{CF :}}$ The auxiliary equation is given by m²+2m+1 =0

⇒ (m+1)² = 0.

⇒ m = -1, -1 (real and equal roots)

∴ CF = (A+Bx)e^-x where A and B are arbitrary constants.

$\boxed{\text{PI :}}$ The particular integral is given by

$\dfrac{1}{D^2+2D+1}e^{-x}$

= $\dfrac{1}{(D+1)^2}e^{-x}$

= $\dfrac{1}{(D+1)^2}e^{-x}$

= $\dfrac{x^2}{2}e^{-x}$

Therefore, the complete solution of (D²+2D+1)y = e^-x is given by y = CF + PI = (A+Bx)e^-x + $\dfrac{x^2}{2}e^{-x}$ where A and B are arbitrary constants.

Also Read: How to Find Integrating Factor [with Solved Examples]

Formula 2:

Particular Integral of xⁿ

The particular integral formula of xn is given as follows: $\dfrac{1}{f(D)} x^{n} = [f(D)]^{-1}x^n$ The following binomial formulas are useful in order to compute [f(D)]-1. 1. (1-x)-1 = 1+x+x2+x3+… 2. (1+x)-1 = 1-x+x2-x3+… 3. (1-x)-2 = 1+2x+3x2+4x3+… 4. (1+x)-2 = 1-2x+3x2-4x3+…

Question 3: Solve $\dfrac{d^2y}{dx^2}+4y=1+x^2$.

Answer:

The equation is (D²+4)y = 1+x².

The complete solution is given by

y = CF + PI

$\boxed{\text{CF :}}$ The auxiliary equation is given by m²+4 =0

⇒ m² = -4.

⇒ m = ±2i (complex roots)

∴ CF = e^0.x(A cos2x + B sin2x) = A cos2x + B sin2x.

$\boxed{\text{PI :}}$ The particular integral is given by

$\dfrac{1}{D^2+4}(1+x^2)$

= $\dfrac{1}{4(1+\frac{D^2}{4})}(1+x^2)$

= $\dfrac{1}{4} (1+\frac{D^2}{4})^{-1}(1+x^2)$

= $\dfrac{1}{4} (1-\frac{D^2}{4}+\frac{D^4}{16}- \cdots)(1+x^2)$

= $\dfrac{1}{4} [1+x^2-\frac{1}{4}(0+2)+0-0+ \cdots)]$

= $\dfrac{1}{4} (1+x^2-\frac{1}{2})$

= $\dfrac{x^2}{4} +\dfrac{1}{8}$

Therefore, the complete solution of (D²+4)y = 1+x² is given by y = CF + PI = A cos2x + B sin2x + $\dfrac{x^2}{4} +\dfrac{1}{8}$ where A and B are arbitrary constants.