I have this equation:
$$x(dx-dy) + y(dx+dy) = 0 $$
I tried to solve it by turning it to fraction-type:
$$\frac{dy}{dx} = \frac{x+y}{x-y}$$
However, I realized that it's not homogeneous, and now I am stuck.
Please help!
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3Hint: right hand side: $\frac{1+y/x}{1-y/x}$. Let $u=y/x, \frac{dy}{dx}=\frac{du}{dx}x+u$. – KittyL Feb 22 '15 at 11:52
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@KittyL thanks! it can then be reduced to separable variables,is there any other standard method for solving this particular type? – Samuel Feb 22 '15 at 11:59
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I didn't see any other way to solve the original equation. This substitution rule is typical. It can be used when the right hand side of $\frac{dy}{dx}=f(x,y)$ can be written as a function of $y/x$. – KittyL Feb 22 '15 at 12:11
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@KittyL got you! – Samuel Feb 22 '15 at 12:15
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if you know a little bit of linear algebra, you will find the eigenvalues of $\pmatrix{1&-1\1&1}$ to be $1\pm i$ and $e^{At} = e^t\pmatrix{\cos t & -\sin t\\sin t & \cos t}$ for the $\frac{d}{dt} (x,y)^T = A(x,y)^T.$ – abel Feb 22 '15 at 15:15
