Not really, is this a rhetorical question? lol
Answer:
y = (11x + 13)e^(-4x-4)
Step-by-step explanation:
Given y'' + 8y' + 16 = 0
The auxiliary equation to the differential equation is:
m² + 8m + 16 = 0
Factorizing this, we have
(m + 4)² = 0
m = -4 twice
The complimentary solution is
y_c = (C1 + C2x)e^(-4x)
Using the initial conditions
y(-1) = 2
2 = (C1 -C2) e^4
C1 - C2 = 2e^(-4).................................(1)
y'(-1) = 3
y'_c = -4(C1 + C2x)e^(-4x) + C2e^(-4x)
3 = -4(C1 - C2)e^4 + C2e^4
-4C1 + 5C2 = 3e^(-4)..............................(2)
Solving (1) and (2) simultaneously, we have
From (1)
C1 = 2e^(-4) + C2
Using this in (2)
-4[2e^(-4) + C2] + 5C2 = 3e^(-4)
C2 = 11e^(-4)
C1 = 2e^(-4) + 11e^(-4)
= 13e^(-4)
The general solution is now
y = [13e^(-4) + 11xe^(-4)]e^(-4x)
= (11x + 13)e^(-4x-4)
The answer is option three(photo for step by step)
Answer:
Cubic polynomial has zeros at x=−1x=−1 and 22, is tangent to x−x−axis at x=−1x=−1, and passes through the point (0,−6)(0,−6).
So cubic polynomial has double zero at x=−1x=−1, and single zero at x=2x=2
f(x)=a(x+1)2(x−2)f(x)=a(x+1)2(x−2)
f(0)=−6f(0)=−6
a(1)(−2)=−6a(1)(−2)=−6
a=3a=3
f(x)=3(x+1)2(x−2)f(x)=3(x+1)2(x−2)
f(x)=3x3−9x−6
