Question

How many solution(s) does equation Dny+a1Dn-1y+a2Dn-2y+...+any=R(x) have?
a) n b) one
c)  two d)  none​

Answer 1

The equation $\mathsf{D^{n}y+a_{1}D^{n-1}y+a_{2}D^{n-2}y+...+a_{n}y=R(x)}$ has $\mathsf{n}$ number of solutions.

Correct option: a) $\mathsf{n}$

Step-by-step explanation:

The given equation is

$\quad \mathsf{D^{n}y+a_{1}D^{n-1}y+a_{2}D^{n-2}y+...+a_{n}y=R(x)}$

$\mathsf{\Rightarrow\dfrac{d^{n}y}{dx^{n}}+a_{1}\dfrac{d^{n-1}y}{dx^{n-1}}+a_{2}\dfrac{d^{n-2}y}{dx^{n-2}}+...+a_{n-1}\dfrac{dy}{dx}+a_{n}y=R(x)}$

This is the general form of a linear equation of $\mathsf{n-th}$ order.

We must remember, $\mathsf{a_{1},a_{2},...,a_{n}}$ are constants or functions of $\mathsf{x}$.

However this equation has $\mathsf{n}$ number of solutions.

Its general solution is given by

$\quad \boxed{\mathsf{y=c_{1}y_{1}+c_{2}y_{2}+...+c_{n}y_{n}+y_{p}}}$

where $\mathsf{y_{1}(x),y_{2}(x),...,y_{n}(x)}$ is a set of $\mathsf{n}$ linearly independent solutions and $\mathsf{c_{1},c_{2},...,c_{n}}$ are arbitrary constants.

Also, $\mathsf{y_{p}}$ is a particular solution.