Is there a closed form for any function f(x,y) satisfying: dfdx+dfdy=xy

Name: Is there a closed form for any function f(x,y) satisfying: dfdx+dfdy=xy
Uploaded: 2022-02-23T17:22:57+00:00
Description: Is there a closed form for any function f(x,y) satisfying: dfdx+dfdy=xy

Forgive for using Mathematica, but the answer are: f(x,y)=16(3x2y−x3)+F(y−x) with F - arbitrary continuously differentiable function

Solution: f(x,y)=−x3+3x2y+6c1(y−x)6 (1)

Forgive for using Mathematica, but the answer are: f(x,y)=16(3x2y−x3)+F(y−x) with F - arbitrary continuously differentiable function

Solution: f(x,y)=−x3+3x2y+6c1(y−x)6 (1)

"Is there a closed form for any function..." was answered by students like you

ajedrezlaproa6j

Answered question

2022-01-18

Is there a closed form for any function f(x,y) satisfying:

\frac{d f}{d x} + \frac{d f}{d y} = x y

Answer & Explanation

Anzante2m

Beginner2022-01-19Added 34 answers

Forgive for using Mathematica, but the answer are:

f (x, y) = \frac{1}{6} (3 x^{2} y - x^{3}) + F (y - x)

with F - arbitrary continuously differentiable function

Charles Benedict

Beginner2022-01-20Added 32 answers

Solution:

f (x, y) = \frac{- x^{3} + 3 x^{2} y + 6 c_{1} (y - x)}{6}

(1)

alenahelenash

Expert2022-01-24Added 556 answers

Is solved by

f (x, y) = x y^{2} / 2 - y^{3} / 6

.One way to get this: Let

f_{x} = a (x, y)

so that

f_{y} = x y - a (x, y)

. Then use commutativity of mixed partials to get a handle on a; we'd have

a_{y} (x, y) = y - a_{x} (x, y)

. Then, why not set

a_{x} = 0

? we get

a_{y} = y

and so

a (x, y) = \frac{y^{2}}{2}

works fine. Since we can solve exact differential equations, we should be able to solve this given that

(f_{x})_{y} = (f_{y})_{x}

, indeed we can, with the above

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