Question

Differential equation

Answer 1

\[xy dx + (x^2-3y)dy=0\]

Answer 2

which topic or methods have you studied so far?

Answer 3

You solve this by turning it into an Exact Diff Equation.
multiply equation by some function, u(y) such that
\[\frac{\delta}{\delta x}(ux^2 -3uy) = \frac{\delta}{\delta y}(uxy)\]
After doing the partial derivatives you get:
\[2ux = ux +u' xy\]
\[u = u'y\]
solve this separable diff equ for u(y)
\[u(y) = y\]
Now our diff equ is Exact:
\[xy^2 + (x^2 y - 3y^2) \frac{dy}{dx} = 0\]
\[\Psi(x,y) = \int\limits xy^2 dx = \frac{1}{2}x^2 y^2 + h(y)\]
\[\frac{\delta}{\delta y}\ \left( \frac{1}{2}x^2 y^2 + h(y) \right) = x^2y - 3y^2\]
\[h(y) = \int\limits -3y^2 dy = -y^3\]
Now we have the general solution
\[\Psi(x,y) = C\]
\[\frac{1}{2} x^2y^2 - y^3 = C\]

Answer 4

For more help with Exact equations go to:
http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx

Answer 5

Perhaps an easier way to work this out if you're not familiar with exact ODEs: Rewrite the ODE as
\[xy\,\mathrm dx+(x^2-3y)\,\mathrm dy=0\iff x\frac{\mathrm dx}{\mathrm dy}+\frac{1}{y}x^2=3\]then substitute \(u=x^2\), so that \(\dfrac{1}{2}\dfrac{\mathrm du}{\mathrm dy}=x\dfrac{\mathrm dx}{\mathrm dy}\) and the ODE can be transformed to
\[\frac{1}{2}\frac{\mathrm du}{\mathrm dy}+\frac{1}{y}u=3\iff \frac{\mathrm du}{\mathrm dy}+\frac{2}{y}u=6\]which is linear in \(u\). An integrating factor would be
\[\mu(y)=\exp\left(2\int\frac{\mathrm dy}{y}\right)=y^2\]We have
\[y^2\frac{\mathrm du}{\mathrm dy}+2yu=\frac{\mathrm d}{\mathrm dy}\left[y^2u\right]=6y^2\implies y^2u=2y^3+C\]Undoing the transformation yields the solution
\[x^2y^2=2y^3+C\]same as the one obtained via treating the ODE as exact.