Question

y′′−8y′+16y=24x+2

Answer 1

Homogeneous differential equation,

y′′−8y′+16y=0

Auxiliary equation,

m²-8m+16=0

(m-4)²=0

m=4,4 (repeated roots)

Therefore, solution is y=(a+bx)e^4x.

Now, particular solution is given by,

$y_p=\frac{1}{(D-4)^2}24x+2\\ =\frac{1}{16}[1-\frac{D}{4}]^{-2}(24x+2)\\ =\frac{1}{16}[1+\frac{D}{2}+3\frac{D^2}{16}+••• ](24x+2)\\ =\frac{1}{16}[24x+2+12]\\ =\frac{1}{16}[24x+14]\\ =\frac{12x+7}{8}$

Thus, solution is $y=(a+bx)e^{4x}+\frac{12x+7}{8}$