The first thing we must do in this case is find the derivatives:
y = a sin (x) + b cos (x)
y '= a cos (x) - b sin (x)
y '' = -a sin (x) - b cos (x)
Substituting the values:
(-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)
We rewrite:
(-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)
sin (x) * (- a-b-7a) + cos (x) * (- b + a-7b) = sin (x)
sin (x) * (- b-8a) + cos (x) * (a-8b) = sin (x)
From here we get the system:
-b-8a = 1
a-8b = 0
Whose solution is:
a = -8 / 65
b = -1 / 65
Answer:
constants a and b are:
a = -8 / 65
b = -1 / 65
rechecking this answer with your lecturer. Tq
Answer:
Step-by-step explanation:
let side of cube=x cm
volume=x³ cm³
again side=(x-3) cm
volume=(x-3)³ cm³
x³-(x-3)³=1385
(a³-b³)=(a-b)(a²+ab+b²)
(x-x+3){x²+x(x-3)+(x-3)²}=1385
3(x^2+x²-3x+x²-6x+9)=1385
3(3x²-9x+9)=1385
9x²-27x+27=1385
9x²-27x+27-1385=0
9x²-27x-1358=0
