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 eax
| 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 = e2x 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 = Aex+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 = e2x is given by y = CF + PI = Aex+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 = e0.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.
Particular Integral of Sinax
| The particular integral formula of sinax is given as follows: $\dfrac{1}{f(D^2)} \sin ax = \dfrac{1}{f(-a^2)} \sin ax$ if f(-a2)≠0 If f(-a2)=0, then the above formula goes as follows: $\boxed{\dfrac{1}{f(D^2)} \sin ax = x \dfrac{1}{\frac{d}{dD}(D^2)} \sin ax}$. The same formula works for cosax. |
| Question 4: Solve $\dfrac{d^2y}{dx^2}+4y=\sin x$. |
Answer:
$\boxed{\text{CF :}}$ As Q3, CF is given by
CF = A cos2x + B sin2x.
$\boxed{\text{PI :}}$ The particular integral is as follows:
$\dfrac{1}{D^2+4}(\sin x)$ [Note here a=1 and f(D2)=D2+4.]
= $\dfrac{1}{f(-1^2)} \sin x$
= $\dfrac{1}{(-1^2+4)} \sin x$
= $\dfrac{\sin x}{3}$
So the solution is y = CF + PI = A cos2x + B sin2x + (sinx)/3 where A and B are arbitrary constants.
| Question 5: Solve $\dfrac{d^2y}{dx^2}+4y=\sin 2x$. |
Answer:
$\boxed{\text{CF :}}$ As Q3, CF is given by
CF = A cos2x + B sin2x.
$\boxed{\text{PI :}}$ The particular integral is as follows:
$\dfrac{1}{D^2+4}(\sin 2x)$ [Here a=1 and f(D2)=D2+4, so f(-22)=-22+4=0.]
= $x \dfrac{1}{\frac{d}{dD}(D^2)} (\sin 2x)$
= $\dfrac{x}{2} \dfrac{1}{D} (\sin 2x)$
= $\dfrac{x}{2} \displaystyle \int (\sin 2x)~dx$
= $\dfrac{x}{2} \dfrac{-\cos 2x}{2}$
= $\dfrac{-x\cos 2x}{4}$.
So the solution is y = CF + PI = A cos2x + B sin2x – (xcos2x)/4.
| Exercise: $\dfrac{d^2y}{dx^2}+4y=\sin^2 x$. |
Particular Integral of Product Function
| The particular integral formula of eaxV is given as follows: $\dfrac{1}{f(D)}$ eaxV = eax $\dfrac{1}{f(D+a)}$ V. |
| Question 6: Solve $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=x^2e^{3x}$. |
Answer:
The equation is (D2-2D+4)y = x2e3x.
The complete solution is given by
y = CF + PI
$\boxed{\text{CF :}}$ The auxiliary equation is given by m2-2m+1 =0
⇒ (m-1)2 = 0.
⇒ m = 1,1 (real, equal roots)
∴ CF = (A+Bx)ex.
$\boxed{\text{PI :}}$ The particular integral is given by
$\dfrac{1}{(D-1)^2}x^2e^{3x}$
= $e^{3x}\dfrac{1}{(D+3-1)^2}x^2$
= $e^{3x}\dfrac{1}{(D+2)^2}x^2$
= $\dfrac{e^{3x}}{4} \left\{\left( 1+\dfrac{D}{2}\right)^{-2} \right\}x^2$
= $\dfrac{e^{3x}}{4} \left\{ 1-2\left(\cdot\dfrac{D}{2}\right)+3 \left(\dfrac{D^2}{4}\right)-\cdots \right\}x^2$
= $\dfrac{e^{3x}}{4} \left(x^2-2x+\dfrac{3}{4}\cdot 2 \right)$
= $\dfrac{e^{3x}}{8} \left(2x^2-4x+3 \right)$
Therefore, the complete solution of (D2+4)y = 1+x2 is given by y = CF + PI = (A+Bx)ex + $\dfrac{e^{3x}}{8} \left(2x^2-4x+3 \right)$ where A and B are arbitrary constants.
Particular Integral of xV
| The particular integral formula of xV is given as follows: $\dfrac{1}{f(D)} xV$ = $x\dfrac{1}{f(D)} V +$ $\left(\dfrac{d}{dD} \dfrac{1}{f(D)} \right)V$. |
This article is written by Dr. Tathagata Mandal, Ph.D in Mathematics from IISER Pune (Algebraic Number Theory), Postdocs at IIT Kanpur & ISI Kolkata. Currently, working as an Assistant Prof. at Adamas University. Thank you for visiting the website.