Answer:
The chevrolet and the honda
Step-by-step explanation:
divide both by 8
Solve -6 ( dy(x))/( dx) + 3 ( d^2 y(x))/( dx^2) + 6 y(x) = e^x sec(x):
The general solution will be the sum of the complementary solution and particular solution.Find the complementary solution by solving 3 ( d^2 y(x))/( dx^2) - 6 ( dy(x))/( dx) + 6 y(x) = 0:
Assume a solution will be proportional to e^(λ x) for some constant λ.Substitute y(x) = e^(λ x) into the differential equation:
3 ( d^2 )/( dx^2)(e^(λ x)) - 6 ( d)/( dx)(e^(λ x)) + 6 e^(λ x) = 0
Substitute ( d^2 )/( dx^2)(e^(λ x)) = λ^2 e^(λ x) and ( d)/( dx)(e^(λ x)) = λ e^(λ x):
3 λ^2 e^(λ x) - 6 λ e^(λ x) + 6 e^(λ x) = 0
Factor out e^(λ x):
(3 λ^2 - 6 λ + 6) e^(λ x) = 0
Since e^(λ x) !=0 for any finite λ, the zeros must come from the polynomial:
3 λ^2 - 6 λ + 6 = 0
Factor:
3 (2 - 2 λ + λ^2) = 0
Solve for λ:
λ = 1 + i or λ = 1 - i
The roots λ = 1 ± i give y_1(x) = c_1 e^((1 + i) x), y_2(x) = c_2 e^((1 - i) x) as solutions, where c_1 and c_2 are arbitrary constants.The general solution is the sum of the above solutions:
y(x) = y_1(x) + y_2(x) = c_1 e^((1 + i) x) + c_2 e^((1 - i) x)
Apply Euler's identity e^(α + i β) = e^α cos(β) + i e^α sin(β):y(x) = c_1 (e^x cos(x) + i e^x sin(x)) + c_2 (e^x cos(x) - i e^x sin(x))
Regroup terms:
y(x) = (c_1 + c_2) e^x cos(x) + i (c_1 - c_2) e^x sin(x)
Redefine c_1 + c_2 as c_1 and i (c_1 - c_2) as c_2, since these are arbitrary constants:
y(x) = c_1 e^x cos(x) + c_2 e^x sin(x)
Determine the particular solution to 3 ( d^2 y(x))/( dx^2) + 6 y(x) - 6 ( dy(x))/( dx) = e^x sec(x) by variation of parameters:
List the basis solutions in y_c(x):
y_(b_1)(x) = e^x cos(x) and y_(b_2)(x) = e^x sin(x)
Compute the Wronskian of y_(b_1)(x) and y_(b_2)(x):
W(x) = left bracketing bar e^x cos(x) | e^x sin(x)
( d)/( dx)(e^x cos(x)) | ( d)/( dx)(e^x sin(x)) right bracketing bar = left bracketing bar e^x cos(x) | e^x sin(x)
e^x cos(x) - e^x sin(x) | e^x cos(x) + e^x sin(x) right bracketing bar = e^(2 x)
Divide the differential equation by the leading term's coefficient 3:
( d^2 y(x))/( dx^2) - 2 ( dy(x))/( dx) + 2 y(x) = 1/3 e^x sec(x)
Let f(x) = 1/3 e^x sec(x):
Let v_1(x) = - integral(f(x) y_(b_2)(x))/(W(x)) dx and v_2(x) = integral(f(x) y_(b_1)(x))/(W(x)) dx:
The particular solution will be given by:
y_p(x) = v_1(x) y_(b_1)(x) + v_2(x) y_(b_2)(x)
Compute v_1(x):
v_1(x) = - integral(tan(x))/3 dx = 1/3 log(cos(x))
Compute v_2(x):
v_2(x) = integral1/3 dx = x/3
The particular solution is thus:
y_p(x) = v_1(x) y_(b_1)(x) + v_2(x) y_(b_2)(x) = 1/3 e^x cos(x) log(cos(x)) + 1/3 e^x x sin(x)
Simplify:
y_p(x) = 1/3 e^x (cos(x) log(cos(x)) + x sin(x))
The general solution is given by:
Answer: y(x) = y_c(x) + y_p(x) = c_1 e^x cos(x) + c_2 e^x sin(x) + 1/3 e^x (cos(x) log(cos(x)) + x sin(x))
Answer:
What? HAHAHAHAHA it's the same. I mean the choices are the same
I love these. It's often called the Shoelace Formula. It actually works for the area of any 2D polygon.
We can derive it by first imagining our triangle in the first quadrant, one vertex at the origin, one at (a,b), one at (c,d), with (0,0),(a,b),(c,d) in counterclockwise order.
Our triangle is inscribed in the 
rectangle. There are three right triangles in that rectangle that aren't part of our triangle. When we subtract the area of the right triangles from the area of the rectangle we're left with the area S of our triangle.
That's the cross product in the purest form. When we're away from the origin, a arbitrary triangle with vertices 
will have the same area as one whose vertex C is translated to the origin.
We set 
That's a perfectly useful formula right there. But it's usually multiplied out:
That's the usual form, the sum of cross products. Let's line up our numbers to make it easier.
(1, 2), (3, 4), (−7, 7)
(−7, 7),(1, 2), (3, 4),
[tex]A = \frac 1 2 ( 1(7)-2(-7) + 3(2)-4(1) + -7(4) - (7)(3)
