[amp_mcq option1=”\\[{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + 2{\\text{xy}} + 6\\]” option2=”\\[{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + 5\\]” option3=”\\[{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + {{\\text{x}}^2}{\\text{y}} + {\\text{x}}{{\\text{y}}^2} + {\\text{xy}}\\]” option4=”\\[{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + {{\\text{x}}^2}{\\text{y}} + {\\text{x}}{{\\text{y}}^2}\\]” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\text{A. }{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + 2{\\text{xy}} + 6}$.<\/p>\n
To find a function $V(x, y)$ that satisfies the given equations, we can use the method of undetermined coefficients. We assume that $V(x, y)$ is a linear combination of monomials of the form $x^n y^m$, where $n$ and $m$ are non-negative integers. Then, the partial derivatives of $V$ with respect to $x$ and $y$ can be written as follows:<\/p>\n
$$\\begin{align} $$\\begin{align} Comparing the coefficients of $x^n y^m$ in the equations $\\frac{\\partial V}{\\partial x} = p x^2 + y^2 + 2xy$ and $\\frac{\\partial V}{\\partial y} = x^2 + q y^2 + 2xy$, we find that the coefficients of $x^n y^m$ must satisfy the following equations:<\/p>\n $$\\begin{align} Solving these equations, we find that $n=p$ and $m=q$. Therefore, the function $V(x, y)$ that satisfies the given equations is<\/p>\n $$V(x, y) = \\sum_{n=0}^{\\infty} (p+q) x^n y^n = \\frac{p}{3} x^3 + \\frac{q}{3} y^3 + 2xy.$$<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”\\[{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + 2{\\text{xy}} + 6\\]” option2=”\\[{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + 5\\]” option3=”\\[{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + {{\\text{x}}^2}{\\text{y}} + {\\text{x}}{{\\text{y}}^2} + {\\text{xy}}\\]” option4=”\\[{\\text{p}}\\frac{{{{\\text{x}}^3}}}{3} + {\\text{q}}\\frac{{{{\\text{y}}^3}}}{3} + {{\\text{x}}^2}{\\text{y}} + {\\text{x}}{{\\text{y}}^2}\\]” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20220","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n
\n\\frac{\\partial V}{\\partial x} &= \\sum_{n=0}^{\\infty} n x^{n-1} y^m \\left( \\frac{d}{dx} x^n \\right) + \\sum_{n=0}^{\\infty} \\left( \\frac{d}{dx} y^m \\right) x^n y^{m-1} \\
\n&= \\sum_{n=1}^{\\infty} n x^{n-1} y^m + \\sum_{m=1}^{\\infty} m y^{m-1} x^n \\
\n&= \\sum_{n=1}^{\\infty} n x^n y^m + \\sum_{m=1}^{\\infty} m x^n y^{m-1} \\
\n&= \\sum_{n=0}^{\\infty} (n+m) x^n y^m,
\n\\end{align<\/em>}$$
\nand<\/p>\n
\n\\frac{\\partial V}{\\partial y} &= \\sum_{n=0}^{\\infty} n x^n y^{m-1} \\left( \\frac{d}{dy} y^m \\right) + \\sum_{n=0}^{\\infty} y^m \\left( \\frac{d}{dy} x^n \\right) \\
\n&= \\sum_{n=0}^{\\infty} n x^n y^{m-1} m + \\sum_{n=0}^{\\infty} x^n y^{m-1} n \\
\n&= \\sum_{n=0}^{\\infty} n x^n y^{m-1} m + \\sum_{n=0}^{\\infty} n x^n y^{m-1} \\
\n&= \\sum_{n=0}^{\\infty} (n+m) x^n y^m.
\n\\end{align<\/em>}$$<\/p>\n
\np &= n+m, \\
\nq &= n+m, \\
\n2 &= n+m.
\n\\end{align<\/em>}$$<\/p>\n