Question

(x^2-y^2 )dx+(y^2-2xy)dy=0

Answer 1

what is the general solution to this

Answer 2

Have you learned about exact equations? I wouldn't want to delve into something you're not at least somewhat familiar with.

Answer 3

Uh, okay. I'll just go ahead with the exact equations method.
\[x^2-y^2+\left(y^2-2xy\right)\frac{dy}{dx}=0\]
For the equation to be exact, you must have
\[\frac{\partial}{\partial x}\left[y^2-2xy\right]=\frac{\partial}{\partial y}\left[x^2-y^2\right]\]
Indeed, you have
\[-2y=-2y\]
and thus the equation is exact.

Answer 4

The general solution will be of the form \(\Psi(x,y)=C\), since by the chain rule we have
\[\begin{align*}\frac{d}{dy}\Psi(x,y)&=\frac{d}{dy}C\\
\Psi_x+\Psi_y\frac{dy}{dx}&=0
\end{align*}\]
where \(\Psi_x\) and \(\Psi_y\) denote the corresponding partial derivatives.

You have \(\Psi_x=x^2-y^2\), which tells you that
\[\begin{align*}\int\Psi_x~dx&=\int\left(x^2-y^2\right)~dx\\
\Psi&=\frac{1}{3}x^3-xy^2+h(y)
\end{align*}\]
Differentiating with respect to \(y\), you have
\[\begin{align*}\Psi_y&=-2xy+\frac{dh}{dy}\\
y^2-2xy&=-2xy+\frac{dh}{dy}\\
y^2~dy&=dh\\
\int y^2~dy&=\int dh\\
\frac{1}{3}y^3+C&=h(y)\end{align*}\]
Thus, the solutions is
\[\Psi=\frac{1}{3}x^3-xy^2+\frac{1}{3}y^3+C~~\Rightarrow~~\frac{1}{3}x^3-xy^2+\frac{1}{3}y^3=C^*\]

Answer 5

Is the equation second order and linear?

Answer 6

Second order? No, the highest-order derivative is one. Linear? Also no, there's a factor of \(y^2\) that makes it non-linear.

Answer 7

do you know how to do this? Verify that y(x)=-1/14+12e^14x is a solution of the ordinary differential equation y^'=14y+1. Note you do not need to solve the ordinary differential equation.

Answer 8

Plug in the given \(y\) into the equation. If equality holds, then it's a solution.

Answer 9

so I plug the top into the bottom one?

Answer 10

Yes.
\[y=-\frac{1}{14}+12e^{14x}~~\Rightarrow~~y'=168e^{14x}\]
Plug these in

Answer 11

That's all I have to do?
so it is verified?