Question

general solution of higher order differential equation depends on​

Answer 1

Answer:

The general form of such an equation is. a0(x)y(n) + a1(x)y(n-1) + ··· + an-1(x)y. /

Answer 2

Answer:

General solution of higher order differential equation depends on its characteristic equation.

Step-by-step explanation:

For a 2nd order differential equation:

$ay'' +by'+cy = 0$

The characteristic equation is $a\lambda^2+b\lambda +c =0$  (Where λ is some constant)

Case1 : If characteristic equation has distinct real roots λ₁ and λ₂, then the solution of differential equation is:

                     $y(x) = c_1e^{\lambda_1 x}+c_2e^{\lambda_2 x}$  (where c₁ and c₂ are constants)

Case2 : If characteristic equation has equal root λ then the solution of differential equation is:

                      $y(x) = c_1e^{\lambda x}+c_2xe^{\lambda x}$

Case3 : If characteristic equation has imaginary roots $\lambda_1 = \alpha +\beta i$ and $\lambda_2 = \alpha -\beta i$, then the solution of differential equation is:

                      $y(x) = c_1e^{(\alpha +\beta i)x} + c_2e^{(\alpha -\beta i)x}$

Similarly, for nth order of differential equation

                  ${y^{\left( n \right)}}\left( x \right) + {a_1}{y^{\left( {n - 1} \right)}}\left( x \right) + \cdots+ {a_{n - 1}}y'\left( x \right) + {a_n}y\left( x \right) = 0$

The characteristic equation is:

                  $L\left( \lambda \right) = {\lambda ^n} + {a_1}{\lambda ^{n - 1}} + \cdots + {a_{n - 1}}\lambda + {a_n} = 0$

Therefore, general solution of higher order differential equation depends on​ characteristic equation.

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