The ODE is $y'(x) + \frac{y(x)}{x}=-x^4y(x)^3$.
I found the solution to be:
$$y(x) = \frac{1}{x\sqrt{\frac{2}{3}x^3+C}}$$
but I'm not sure what is meant by "singular" solutions.
Thanks!
The ODE is $y'(x) + \frac{y(x)}{x}=-x^4y(x)^3$.
I found the solution to be:
$$y(x) = \frac{1}{x\sqrt{\frac{2}{3}x^3+C}}$$
but I'm not sure what is meant by "singular" solutions.
Thanks!
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It is a solution to the differential equation that isn't actually included in the solution you found above if such a solution exists. For example, your solution says y cannot be 0. But y can be 0. Since 0+0=0
The singular solution is $y=0$