# Linear Differential Equations

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 form$x^n \dfrac{d^n y}{dx^n} + p_1 x^{n- 1} \dfrac{d^{n- 1}}{dx^{n- 1}} + \cdots + p_{n- 1} x \dfrac{dy}{dx} + p_n y = Q$ 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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