Solving the Bernoulli Differential Equation x^2(dy/dx) + y^2 = xy

How to solve a Bernoulli Differential Equation

[angelmendez-rivera]

This equation is actually a better example of a homogeneous equation. Given y' – y/Id = –(y/Id)^2, the better substitution here is z = y/Id, equivalent to Id·z, hence Id·z' + z = y', hence y' – z = Id·z'. Thus, Id·z' = –z^2, and this is separable. Hence z = 0, or –z'/z^2 = 1/Id, hence 1/z = ln(–Id) + A if Id < 0, 1/z = ln(Id) + B if Id > 0, so z = 1/[ln(–Id) + A] if Id < 0, z = 1/[ln(Id) + B] if Id > 0, hence y = Id/[ln(–Id) + A] if Id < 0, y = Id/[ln(Id) + B] if Id > 0. This definitely works better using initial conditions, though. If –z'/z^2 = 1/Id. Integrate over [x, –1] and over [1, x], respectively. We have that 1/z(–1) – 1/z(x) = ln(–1/x) for x < 0 and 1/z(x) – 1/z(1) = ln(x) for x > 0, which translates to 1/z(x) = 1/z(–1) + ln(–x) for x < 0 and 1/z(x) = 1/z(1) + ln(x) for x > 0. Hence, for every x, y(x) = 0, or, for every x < 0, y(x) = x/[ln(–x) – 1/y(–1)], and for every x > 0, y(x) = x/[ln(x) + 1/y(1)]. More concretely, what we have is, for every x < 0, y(x) = y(–1)·x/[y(–1)·ln(–x) – 1], and for every x > 0, y(x) = y(1)·x/[y(1)·ln(x) + 1], or for every x, y(x) = 0. Even more succintly, for every x, y(x) = 0, or y(x) = + sgn(x)}. As for the latter, notice that lim y (x —> 0) = 0. Since y' – y/Id = –(y/Id)^2, we have that lim y' – y/Id (x —> 0) = lim –(y/Id)^2 (x —> 0), and since lim y/Id (x —> 0) = 0, we have that lim y' (x —> 0) = 0. So, if we define y(0) = 0, then for every nonzero real x, y(0) = 0, y(x) = + sgn(x)}, and with y defined in this fashion for every real x, y is continuously differentiable everywhere, and includes y(x) = 0 in the case that y[sgn(x)] = 0. This is what thr video misses.

[saikiranyalla]

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[carlossilva-hlje]

THANKS MAN, NICE VIDEO GREETING YOU FROM MEXICO

[FernandoGomez-hgrn]

Awesome video, thanks for the detailed solution.

[PandazHub]

Hows it feel to know you save lives? :)))

[dragosmanailoiu]

much easier to use homogeneous substitution here

[saikiranyalla]

Solution of the equation y'+y= e^x^x is?