exact - 4

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle xy \ dx + \left(2x^2 + 3y^2 - 20\right) \ dy = 0\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle xy \ dx + \left(2x^2 + 3y^2 - 20\right) \ dy = 0\)

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

\(\displaystyle xy \ dx + \left(2x^2 + 3y^2 - 20\right) \ dy = 0\)

Let

\(\displaystyle M = xy\)

and

\(\displaystyle N = 2x^2 + 3y^2 - 20\)


\(\displaystyle \frac{\partial M}{\partial y} = x\)


\(\displaystyle \frac{\partial N}{\partial x} = 4x\)


Since \(\displaystyle \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), we have to do more work to solve this problem.

 

[logistic_guy]

The trick is to have an integrating factor.

\(\displaystyle \frac{N_x - M_y}{M} = \frac{4x - x}{xy} = \frac{3x}{xy} = \frac{3}{y}\)

Then, the integrating factor is:

\(\displaystyle \mu(y) = e^{\int\frac{3}{y}dy} = y^3\)

Our differential equation becomes:

\(\displaystyle xy^4 \ dx + \left(2x^2y^3 + 3y^5 - 20y^3\right) \ dy = 0\)

 

[logistic_guy]

\(\displaystyle xy^4 \ dx + \left(2x^2y^3 + 3y^5 - 20y^3\right) \ dy = 0\)

Let

\(\displaystyle M = xy^4\)

And

\(\displaystyle N = 2x^2y^3 + 3y^5 - 20y^3\)

Since we have:

\(\displaystyle \frac{\partial M}{\partial y} = 4xy^3 = \frac{\partial N}{\partial x}\)

The differential equation is exact and we can solve it.

 

[logistic_guy]

We continue like pros.

\(\displaystyle \frac{\partial f}{\partial x} = M = xy^4\)


\(\displaystyle \int \frac{\partial f}{\partial x} \ dx = \int xy^4 \ dx\)


\(\displaystyle f(x,y) = \frac{x^2y^4}{2} + g(y)\)


\(\displaystyle \frac{\partial f}{\partial y} = 2x^2y^3 + g'(y)\)


\(\displaystyle \frac{\partial f}{\partial y} = N = 2x^2y^3 + g'(y)\)


\(\displaystyle 2x^2y^3 + 3y^5 - 20y^3 = 2x^2y^3 + g'(y)\)


\(\displaystyle 3y^5 - 20y^3 = g'(y)\)


\(\displaystyle \int (3y^5 - 20y^3) \ dy = \int g'(y) \ dy\)


\(\displaystyle \frac{y^6}{2} - 5y^4 = g(y)\)

This gives:

\(\displaystyle f(x,y) = \frac{x^2y^4}{2} + \frac{y^6}{2} - 5y^4\)

Then the general solution to the differential equation is:

\(\displaystyle \frac{x^2y^4}{2} + \frac{y^6}{2} - 5y^4 = D\)

Or

\(\displaystyle x^2y^4 + y^6 - 10y^4 = C\)

Or

\(\displaystyle (x^2 - 10)y^4 + y^6 = C\)