answer:Consider the differential equation {eq}{y}'' + 4y' + 13y = 0cdots cdots (1);, , y(0)=2,: : : y'(0)=-3 L left { f^n right }=s^nF(s)-s^{n-1}f(0)-s^{n-2}f^{'}(0)-cdots -sf^{n-2}(0)-f^{n-1}(0)cdots cdots (A) {/eq} Taking Laplace (1) on both sides {eq}Lleft { y^{''} right }+4Lleft { y^{'} right }+13Lleft { y right }=0 {/eq} Using formula (A) {eq}s^{2}Y(s)-sy(0)-y{}'(0)+4sY(s)-4y(0)+13Y(s)=0 s^{2}Y(s)-2s+3+4sY(s)-8+13Y(s)=0;: : : : left ( y(0)=2,: : : y'(0)=-3 right ) s^{2}Y(s)+4sY(s)+13Y(s)=5 (s^{2}+4s+13)Y(s)=5 Y(s)=frac{5}{(s^{2}+4s+13)} {/eq} Taking laplace inverse on both sides {eq}y(t)=L^{-1}left {frac{5}{(s^{2}+4s+13)} right } =5L^{-1}left{frac{1}{(s+2)^2+9}right} =frac{5}{3}e^{-2t}sin left(3tright) {/eq}

answer:The projection of v_1 onto v_2 is calculated as follows: 1. Compute the dot product of v_1 and v_2: v_1 cdot v_2 = left(-1right) left(frac{3}{2}right) + 0 left(-frac{1}{2}right) + left(-frac{3}{2}right) left(frac{1}{2}right) + left(frac{5}{2}right) left(-frac{3}{2}right) + left(frac{1}{2}right) left(-frac{1}{2}right) + left(frac{5}{2}right) left(-1right) = -frac{21}{4} 2. Calculate the magnitude of v_2: ||v_2|| = sqrt{left(frac{3}{2}right)^2 + left(-frac{1}{2}right)^2 + left(frac{1}{2}right)^2 + left(-frac{3}{2}right)^2 + left(-frac{1}{2}right)^2 + (-1)^2} = sqrt{frac{31}{4}} 3. Find the projection coefficient: text{Coefficient} = frac{v_1 cdot v_2}{||v_2||^2} = frac{-frac{21}{4}}{frac{31}{4}} = -frac{21}{31} 4. Project v_1 onto v_2: text{Projection of } v_1 text{ onto } v_2 = -frac{21}{31} cdot v_2 = left{-frac{21}{31} cdot frac{3}{2}, -frac{21}{31} cdot left(-frac{1}{2}right), -frac{21}{31} cdot frac{1}{2}, -frac{21}{31} cdot left(-frac{3}{2}right), -frac{21}{31} cdot left(-frac{1}{2}right), -frac{21}{31} cdot (-1)right} = left{-frac{63}{62}, frac{21}{62}, -frac{21}{62}, frac{63}{62}, frac{21}{62}, frac{21}{31}right} However, upon further verification, it seems there was an error in the original answer due to incorrect scaling. The correct answer after performing the calculations is:

answer:The curl of the vector field vec{F} is given by the cross product of the gradient of each component with the unit vector in the direction of the other two coordinates. The curl is calculated as follows: nabla times vec{F} = left( frac{partial}{partial y} h(x, y, z) - frac{partial}{partial z} g(x, y, z) right)uvec{i} - left( frac{partial}{partial x} h(x, y, z) - frac{partial}{partial z} f(x, y, z) right)uvec{j} + left( frac{partial}{partial x} g(x, y, z) - frac{partial}{partial y} f(x, y, z) right)uvec{k} Let's compute each component: 1. frac{partial}{partial y} h(x, y, z) - frac{partial}{partial z} g(x, y, z) = 5y^4z^5 - 0 = 5y^4z^5 2. frac{partial}{partial x} h(x, y, z) - frac{partial}{partial z} f(x, y, z) = 0 - frac{5left(x^5+yright)^4 cdot 5x^4}{z^6} = -frac{25x^4 left(x^5+yright)^4}{z^6} 3. frac{partial}{partial x} g(x, y, z) - frac{partial}{partial y} f(x, y, z) = 0 - frac{5 left(x^5+yright)^4}{z^5} = -frac{5 left(x^5+yright)^4}{z^5} So the curl of the vector field is: nabla times vec{F} = left{5y^4z^5, -frac{25x^4 left(x^5+yright)^4}{z^6}, -frac{5 left(x^5+yright)^4}{z^5}right}

answer:To solve this quadratic equation, we can use the factoring method. The coefficients of x and the constant term are 12 and 32, respectively. We seek two numbers that multiply to 32 and add up to 12. These two numbers are 8 and 4 because: 8 times 4 = 32 8 + 4 = 12 Hence, the equation can be factored as: (x + 8)(x + 4) = 0 By the zero-product property, if the product of two terms is zero, then one or both of the terms must be zero. Therefore, we set each factor equal to zero and solve for x: x + 8 = 0 quad text{or} quad x + 4 = 0 Solving for x gives us the following solutions: x = -8 quad text{and} quad x = -4 Thus, the solutions to the equation x^2 + 12x + 32 = 0 are x = -8 and x = -4.