Is My Solution to the Exact Differential Equation Correct?

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The discussion focuses on verifying the correctness of a solution to an exact differential equation. The user outlines their approach, confirming that the equation is exact by showing that the partial derivatives of M and N are equal. They derive the solution and express it in terms of a function F(y), ultimately arriving at the equation x^2y - 3x^3 + y^2 + y = c. The importance of checking the solution using total derivatives is emphasized, and the user successfully manipulates their expression back to the original form. The overall consensus is that the solution and approach are correct and align with textbook methods.
chwala
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Homework Statement
Solve the exact differential equation
Relevant Equations
exact equations
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now my approach is different, i just want to check that my understanding on this is correct.

see my working below;
 
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##2xy-9x^2+(2y+x^2+1)\frac {dy}{dx}=0##
##2xy-9x^2dx+(2y+x^2+1)dy=0##
Let ##M(x,y)=2xy-9x^2##
##N(x,y)=2y+x^2+1## Since ##\frac {∂M}{∂y}=2x=\frac {∂N}{∂x}=## then the differential equation is exact.
Therefore, ##\int Mdx## = ##x^2y-3x^3+F(y)##........1
and ##\int Ndy## = ##y^2+x^2y+y+c ##......2
therefore, ##F(y)= y^2+y+c##......3

therefore, we shall have (from 1 and 3), ## x^2y-3x^3+y^2+y=c##
i understand it this way better, i just want to know if this is also correct.
 
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Looks fine to me.
 
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That's how I understand it better from my undergraduate studies...thanks
 
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chwala said:
i just want to know if this is also correct.
Once you have your solution, it's good practice to check by finding the total derivative of your expression, which you should be able to manipulate back into the form the equation was given in.
 
The solution is correct, its a textbook question...my interest was on the approach or rather my way of working the problem to realize the solution.
Thanks Mark for your input. Yeah I will use total derivatives to check the solution...
 
Mark44 said:
Once you have your solution, it's good practice to check by finding the total derivative of your expression, which you should be able to manipulate back into the form the equation was given in.

just to follow your guidance, on checking...
let ##u=x^2y-3x^3+y^2+y##
##f_{x}=2xy-9x^2##
##f_{y}=x^2+2y+1##
therefore,
##du=f_{x} dx+f_{y} dy##
##du=(2xy-9x^2)dx+(x^2+2y+1)dy##
bingo!
 

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