Linear Differential Equations


Posted in Math Formulas




Linear Differential Equation of nth Order

Linear differential equation is in the form of Ly=f, where ‘L’ is a linear operator, ‘Y’ is a unknown function and ‘F’ is a known function of a same nature. In mathematical point of view first-order linear differential equation are those equation that can be kept in \dfrac{dy}{dx} + p ( x) y = Q (x) form.

1. A differential equation of the form

\dfrac{d^ny}{dx^n} + p_1\dfrac{d^{n- 1}y}{dx^{n- 1}} + p_2 \dfrac{d^{n- 2}}{dx^{n- 2}} + \cdots \cdots + p_ny = Q where p_1, p_2 \cdots \cdots p_n are constant and Q in any function of x or constant is called linear differential equating of nth order with constant co-effieient.

Note:

i. \dfrac{d^2 y}{dx^2} + p_1 \dfrac{dy}{dx} + p_2y = 0 \cdots (1)

ii.\dfrac{d^2y}{dx^2} + p_1 \dfrac{dy}{dx} + p_2y = Q \cdots \cdots (2) are equation of second order.

2.We use \dfrac{d}{dx} = D; \dfrac{dy}{dx} = Dy; \dfrac{d^2y}{dx^2} = D^2y

3.Solution f the equation f (D) y = Q

 

Case 1:

If Q = 0, f_1 (x), y = f_2(x) are independent solutions of the equation (D^2 + p_1D + p_2)y = 0, or f (D)y = 0 then, y = c_1f_1 (x) + c_2f_2 (x) will also be solution of the equation.

 

Case 2:

If Q \neq 0, y = F (x)be a general solution of the equation f (D)Y = 0 and y = \Phi (x) be particular solution of f (D)Y = Q then y = F (x) + \Phi (x) is the general solution of f (D)y = Q.

 

4.If m replace D by m then the equation

(D^2 + p_1D + p_2)y = 0 reduces to m^2 + p_1m + p_2)y = 0 but y = e^{mx} \neq 0

So, m^2 + p_1m + p_2)y = 0 \cdots(i) is called auxiliary equation. Equation (i) is quadratic in m and must have two roots.

 

Case 1: auxiliary equation having different and real roots, equal roots, imaginary roots.

Roots Solution General solution
\alpha \neq \beta y = e^{\alpha x}, y = e^{\beta x} y = c_1e^{\alpha x} + c_2e^{\beta x}
\alpha = \beta y = e^{\alpha x}, y = e^{\alpha x} y = (c_1 + xc_2)e^{\alpha x}
\alpha + i\beta, \alpha- i\beta y = c_1e^{(\alpha + I \beta)x}, y = e^{(\alpha- I \beta)} y = e^{\alpha x}(A \cos \beta x + B \sin \beta x)

 

5. Some useful relations to find P.I.

i. \dfrac{1}{(D- \alpha)}Q = e^{\alpha x} \int Qe^{-\alpha x} dx, \alpha \, \, is \, \, constant \\ ii. \dfrac{1}{f (D)}e^{ax} \, \, when \, \, a \neq 0 \\ iii. \dfrac{1}{f (D)} e^{ax} \, \, when \, \, f (a) = 0 \\ \dfrac{1}{f (D)} e^{ax} = \dfrac{x.e^{ax}}{f^1(a)} = x \dfrac{1}{f^1 (D)} e^{ax} \\ when \, \, f^1(a) = 0 \, \, then \, \, \dfrac{1}{f (D)} e^{ax} = x^2 \dfrac{1}{f^{11}(D)}e^{ax} \\ iv. \dfrac{1}{f(D^2)} \sin ax = \dfrac{1}{f(-a^2)} \sin ax, f(-a^2) \pm 0 \\ \dfrac{1}{f (D^2)}\sin (ax + b) = \dfrac{1}{f(-a^2)}\sin (ax + b) \\ v. \dfrac{1}{f (D^2)} \cos ax = \dfrac{1}{f (-a^2)} \cos ax \\ \dfrac{1}{f (D^2)} \cos (ax + b) = \dfrac{1}{f(-a^2)} \cos (ax + b), f(-a^2) \neq 0
6 . \dfrac{1}{(-a^2)}x^m where m is positive integer

 

In this case we have to use binomial theorem and then find derivatives successively, for this we need following relations.

i. (1 + D)^{-1} = 1- D + D^2 + D^3 + \cdots \\ ii. (1- D)^{-1} = 1 + D + D^2 + D^3 + \cdots \\ iii. (1 + D)^{-2} = 1- 2D + 3D^2- 4D^3 + \cdots \\ iv. (1- D)^{-1} = 1 + 2D + 3D^2 +4 D^3 + \cdots

 

7. \dfrac{1}{f (D)} e^{ax}v , v is function of x or constant.

\dfrac{1}{f (D)} e^{ax}v = e^{ax} \dfrac{1}{f(D + a)} v

 

8. \dfrac{1}{f (D)}xv = x \dfrac{1}{f (D)}v- \dfrac{f'(D)}{(f (D))^2} v v is function of x.

 

9.\dfrac{1}{f (D)} x^m. \cos (ax + b) = \dfrac{1}{f (D)} [=real part of x^m is e^{i(ax + b)}] \\ \dfrac{1}{f (D)}x^m. \sin (ax + b) = \dfrac{1}{f (D)} [co-efficient of I in x^m e^{i (ax + b)}]

 

10. Homogenous Linear equation

 

An equation of the formx^n \dfrac{d^n y}{dx^n} + p_1 x^{n- 1} \dfrac{d^{n- 1}}{dx^{n- 1}} + \cdots + p_{n- 1} x \dfrac{dy}{dx} + p_n y = Q where p_1, p_2, p_n are constant and Q is function of x, is called homogenous linear equation

 

Note:

i. x^n \dfrac{d^n y}{dx^2} + p_1 x\dfrac{dy}{dx} + p_2y = Q is homogeneous linear equation of second order.

ii. e^z = x \\ \delta y = x \dfrac{dy}{dx} \\ \delta (\delta- 1) = x^2 \dfrac{d^2y}{dx^2}

 

11. Equation Reducible to the Homogeneous Linear Form.

 

The equation of the form

(a + bx)^2 \dfrac{d^2y}{dx^2} + p_1 (a + bx) \dfrac{dy}{dx} + p_2y = Q is reduced to the homogeneous linear form by putting;

A + bx = t so that \dfrac{dy}{dx} = \dfrac{dy}{dt}.\dfrac{dt}{dx} = b\dfrac{dy}{dt} \, \, or \, \, \dfrac{d^3y}{dx^2} = b^2 \dfrac{d^2y}{dt^2} the equation can be directly reduced to one with constant co-efficient by putting (a + bx) = e^z \, \, so \, \, that \, \, z = \log (a + bx)



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