## Trending News

# how the equation is not exact(5x^2y+6x^2y^3+4xy^2)dx+(2x^3+3x^4y+3x^2y)dy=0?

### 5 Answers

- 9 years agoFavorite Answer
Consider the equation (5x^2y+6x^3y^2+4xy^2)dx+(2x^3+3x^4y+3x^2y)dy=0 (a) show that the equation is not exact (b) Multiply the equation by x'' and y''' and determine values for n and m that make the resulting equation exact. (c) Use the solution of the resulting exact equation to solve the original equation.

- 7 years ago
a)

(5x^2y+6x^3y^2+4xy^2)dx+(2x^3+3x^4y+3x^2y)dy

∂M/∂y (5x^2y+6x^3y^2+4xy^2) =/= ∂N/∂x (2x^3+3x^4y+3x^2y

5x^2+12x^3y+8xy =/= 6x^2+12x^3y+6xy

b)

Take x^ny^m and multiply it through the equation:

(5x^2y+6x^3y^2+4xy^2)dx+(2x^3+3x^4y+3x^2y)dy X x^ny^m

(5x^2x^nyy^m+6x^3x^ny^2y^m+4xx^ny^2y^m)dx+(2x^3x^ny^m+3x^4x^nyy^m+3x^2x^nyy^m)dy

Now for example, yy^m is the same as y^(m+1) and y^2y^m is y^(m+2). So do this throughout the equation, after; we derive the partial derivative based on these values:

(m+1)5x^(n+2)y^m+(m+2)6x^(n+3)y^(m+1)+(m+2)4x^(n+1)y^(m+1)

= (n+2)2x^(n+2)y^m+(n+4)3x^(x+3)y^(m+1)+(n+2)3x^(n+1)y^(m+1)

Now the coefficients need to be equal, since the x and y powers are equal:

5(m+1) + 6(m+2) + 4(m+2) = 2(n+3) + 3(n+4) + 3(n+3)

15m + 1 = 8n

m = 1, n =2 satisfies this!

c)

Now plug it in to get an exact equation:

(5x^4y^2 + 6x^5y^3 + 4x^3y^3)dx + (2x^5y + 3x^6y^2 + 3x^4y^2)dy

Now solve it like any other exact equation. I'll give the solution I got (doesn't mean my solution is correct);

My Solution: x^5y^2+x^6y^3+x^4y^3+y

- Anonymous9 years ago
if ∂N/∂x =/= ∂M/∂y, then it's not exact

in other words, the partial derivatives of the each of the expressions (namely M & N) with respect to x, & y should be exactly the same, to be considered exact, then you can determine an arbitrary function Ψ.

- How do you think about the answers? You can sign in to vote the answer.
- exyLv 44 years ago
F*cking genuine. i truly love math, algebra, calculus, geometry, and so on. and that i might take a seat and try this each and every at some point if i had to. in fact I even locate new the thank you to choose new formula, and equations. i assume that is why I have been given all "A's" in college