Answer to Question #345432 in Differential Equations for Yuri

Question #345432

(π‘₯𝑦2+ 𝑦 βˆ’ π‘₯)𝑑π‘₯ + π‘₯(π‘₯𝑦 + 1)𝑑𝑦 = 0


1
Expert's answer
2022-05-31T08:47:10-0400
"\\dfrac{\\partial Q}{\\partial x}=2xy+1=\\dfrac{\\partial P}{\\partial y}"

The system of two differential equations that define the functionΒ "u(x, y)" is


"\\begin{cases}\n \\dfrac{\\partial u}{\\partial x}=xy^2+y-x \\\\ \\\\\n \\dfrac{\\partial u}{\\partial y}=x^2y+x \n\\end{cases}"

Integrate the first equation over the variableΒ "x"


"u=\\int(xy^2+y-x)dx+\\varphi(y)"

"=\\dfrac{x^2y^2}{2}+xy-\\dfrac{x^2}{2}+\\varphi(y)"

Differentiate with respect to "y"


"\\dfrac{\\partial u}{\\partial y}=x^2y+x+\\varphi'(y) =x^2y+x"

"\\varphi'(y) =0"

"\\varphi(y)=-\\dfrac{C}{2}"

Then


"u=\\dfrac{x^2y^2}{2}+xy-\\dfrac{x^2}{2}-\\dfrac{C}{2}"

The general solution of the exact differential equation is given by


"x^2y^2+2xy-x^2=C"


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