Is My Solution to the Exact Differential Equation Correct?

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SUMMARY

The discussion confirms the correctness of the solution to the exact differential equation represented by the expression ##2xy-9x^2+(2y+x^2+1)\frac {dy}{dx}=0##. The participants validate that the functions ##M(x,y)=2xy-9x^2## and ##N(x,y)=2y+x^2+1## satisfy the condition for exactness, as the partial derivatives are equal. The integration of these functions leads to the solution ##x^2y-3x^3+y^2+y=c##. Additionally, the practice of verifying the solution through total derivatives is emphasized as a crucial step in confirming the accuracy of the derived expression.

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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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