Answer:
a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
Step-by-step explanation:
for the equation
(1 + x⁴) dy + x*(1 + 4y²) dx = 0
(1 + x⁴) dy = - x*(1 + 4y²) dx
[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx
∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx
now to solve each integral
I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁
I₂= ∫[-x/(1 + x⁴)] dx
for u= x² → du=x*dx
I₂= ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ = - tan⁻¹ (x²) +C₂
then
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C
for y(x=1) = 0
1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C
since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for π*N , we will choose for simplicity N=0 . hen an explicit solution would be
1/2 * 0 = - π/4 + C
C= π/4
therefore
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
Write the word problem in algebraic form:
n2 -14 = 5n
Write as a quadratic:
n2 - 5n - 14 = 0
Factor:
(n+2)(n-7) = 0
Solve:
n+2 = 0 => n = -2
n-7 = 0 => n = 7
positive solution is 7. (14 less than 72 = 35 which is also 5 times 7)
1) scale factor 6 / 2 = 3 or 300%
2) z = three times less than 9 = 3
z=3
f(x) = x³ - 8x² + 22x - 20
given x = a, x = b are zeros of a polynomial then
(x - a) and (x - b) are factors and the polynomial is the product of the factors
f(x) = k(x - a)(x - b) → ( k is a multiplier )
note that complex zeros occur in conjugate pairs
3 + 1 is a zero then 3 - i is a zero
zeros are x = 2, x = 3 + i and x = 3 - i, thus
(x - 2 ),(x - (3 + i )) and (x - (3 - i )) are the factors
f(x) = (x - 2)(x - 3 - i )(x - 3 + i)
= (x - 2)(x² - 6x + 10)
= x³ - 8x² + 22x - 20
The expression 
expands to 
.
Hope this helps.
