2. First show that for the completely insulated bar

ut = c2uxx; 0 < x < L; t > 0:
ux(0; t) = 0; t > 0
ux(L; t) = 0; t > 0
u(x; 0) = f(x); 0 < x < L

the separation of variables gives the solution

u(x; t) = A0 + 1 Ancos n x e ( cn )2t
L

n=1 L
X
where
2 Z0 L n x
An = f(x)cos dx; n = 1; 2; 3:::;

L L
then nd the temperature in the bar with L = ; c = 1; and f(x) = x

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

27/4x^6 y^8

step-by-step explanation:

4(3x^2y^4)^3 (2x^3y^5)^4

=

108x^6 y^12 /16x^12 y^20

=

108 /16x^6 y^8

=

27 /4x^6 y^8

it would be 3 ÷by 6 because the number on each side