Meta

🔍

question:Solve the initial value problem for the differential equation: {eq}frac{dx}{dt} + x = cos t, quad x(0) = 0{/eq}

answer:To solve this first-order linear differential equation, we use the integrating factor method. The given equation is: {eq}frac{dx}{dt} + x = cos t{/eq} The integrating factor is the exponential of the coefficient of ( x ) in the differential equation: {eq}text{Integrating factor} = e^{int 1, dt} = e^t{/eq} Now, multiply the entire equation by the integrating factor: {eq}e^t left(frac{dx}{dt} + xright) = e^t cos t{/eq} This simplifies to: {eq}frac{d}{dt}(xe^t) = e^t cos t{/eq} Now, integrate both sides with respect to ( t ): {eq}xe^t = int e^t cos t, dt{/eq} To solve the integral, use integration by parts: Let ( u = cos t ), ( du = -sin t, dt ) Let ( dv = e^t, dt ), ( v = e^t ) The integral becomes: {eq}xe^t = e^t sin t + int e^t sin t, dt{/eq} Now, apply integration by parts again for the remaining integral: {eq}xe^t = e^t sin t - (e^t cos t - int e^t cos t, dt){/eq} Simplify the expression: {eq}xe^t = frac{2e^t sin t - e^t cos t}{2}{/eq} Finally, apply the initial condition ( x(0) = 0 ) to find the constant: {eq}0 = frac{2(0) - (1)}{2} Rightarrow C = 0{/eq} Therefore, the solution to the initial value problem is: {eq}x(t) = frac{e^t (sin t - cos t)}{2}{/eq}

question:Find the Jacobian matrix of the vector-valued function mathbf{F}(x, y, z) = begin{pmatrix} f(x,y,z) g(x,y,z) h(x,y,z) end{pmatrix}, where f(x,y,z) = x^3, g(x,y,z) = frac{1}{z^8}, and h(x,y,z) = cos(z^4).

answer:The Jacobian matrix of mathbf{F} is given by: frac{partial mathbf{F}}{partial mathbf{x}} = left( begin{array}{ccc} frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z} frac{partial g}{partial x} & frac{partial g}{partial y} & frac{partial g}{partial z} frac{partial h}{partial x} & frac{partial h}{partial y} & frac{partial h}{partial z} end{array} right) = left( begin{array}{ccc} 3x^2 & 0 & 0 0 & 0 & -frac{8}{z^9} 0 & 0 & -4z^3sin(z^4) end{array} right)

question:Multiply the following two matrices: left( begin{array}{ccccc} 1 & frac{27}{16} & -frac{11}{8} & frac{1}{16} & -frac{7}{8} -frac{15}{8} & frac{45}{16} & frac{37}{16} & -frac{11}{16} & frac{45}{16} frac{5}{4} & -frac{15}{8} & -frac{15}{16} & -frac{3}{8} & frac{21}{16} end{array} right) and left( begin{array}{c} frac{9}{4} frac{17}{8} frac{23}{16} -frac{17}{8} -frac{23}{8} end{array} right)

answer:The result of the matrix multiplication is: left( begin{array}{c} frac{799}{128} -frac{395}{256} -frac{1407}{256} end{array} right)

question:Rewrite the equation 2x + 3y = 6 into slope-intercept form.

answer:To rewrite the equation 2x + 3y = 6 into slope-intercept form, we need to isolate y on one side of the equation. First, we subtract 2x from both sides of the equation: 2x + 3y - 2x = 6 - 2x Simplifying: 3y = -2x + 6 Next, we divide both sides of the equation by 3: 3y/3 = (-2x + 6)/3 Simplifying: y = (-2/3)x + 2 This is the equation in slope-intercept form, where the slope is -2/3 and the y-intercept is 2.