Question

14 Particular integral of differential equation (D^2 – 9)y = e3x + 1 is​

Answer 1

Step-by-step explanation:

(D^2 - 9)y= e^3x + 1

y=1/(D^2 - 9) * (e^3x + 1)

derivate wrt to D in denominator and multiply by x in numerator

y={x(e^3x + 1)}/ 2D

put D = 3

y={x(e^3x + 1)}/6

Hence P.I. = {x(e^3x + 1)}/6

Answer 2

Given,

equation: $(D^{2}-9)y=e^{3x} +1$

To Find,

particular integral of the given equation.

Solution,

equation given is $(D^{2}-9)y=e^{3x} +1$

$y=\frac{1}{D^{2}-9 } (e^{3x}+1)$

differentiate w.r.t D in the denominator and multiply by x in the numerator

$y=\frac{[x(e^{3x}+1)] }{2D}\\ put D=3\\y=\frac{[x(e^{3x}+1)] }{6}$

Hence particular integral of the equation $(D^{2}-9)y=e^{3x} +1$ is $\frac{[x(e^{3x}+1)] }{6}$.