# Negative square root solution to Bernoulli equation

Find the solution to the following Bernoulli equation subject to the given boundary condition $$\frac{dy}{dx} + \frac{y}{x} = \frac{3}{2y}$$ $y=-1$ at $x=1$ New dependent variable $z=y^2$. Then $$\frac{dz}{dx}=2y\frac{dy}{dx}$$ and the equation becomes $$\frac{dz}{dx} + 2\frac{z}{x} = 3$$ which is linear. Integrating factor $R(x) = x^2$ then $x^2z = x^3 + C$ --> $z = y^2 = x + \frac{C}{x^2}$
Imposing $y=-1$ at $x=1$ gives $C=0$ and $y=-\sqrt{x}$

Why is the answer negative square root of x rather than positive?

Knowing that $y^2 = x +\frac C{x^2}$ and $y(1)=-1$ first gives you that $(-1)^2 = 1 + C$ so $C=0$. Therefore, you know that $$y^2 = x.$$ so either $y(x) = \sqrt x$ or $y(x) = -\sqrt x$. In the first option, you have $y(1) = \sqrt 1 = 1$ which is not correct, in the second, you have $y(1) = -\sqrt 1 = -1$ which is correct.
If you do not use the boundary condition, the formal solutions of the differential equation are $$y=\pm \frac{\sqrt{c_1+x^3}}{x}$$ and, as explained by 5xum, it is the condition which removes one of them.