I have this problem, $\frac{dy}{dx}=\frac{2x}y$, $y(6)=-4$ where the solution $y(x)=\pm\sqrt{2x^2+C}$, $C=-56$, is shown, however I don't understand how the correct sign $y(x)=-\sqrt{2x^2-56}$ is found for the square root. square rooting both sides causes the right side to be plus or minus, and I need to find which one is correct.
Finding correct sign for a square root in a differentiable equation
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ordinary-differential-equations
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0When you substitute in the initial condition, you have that $-4$ equals plus or minus the square root of some quantity. In order for both quantities to be equal, they must also have the same sign. Since -4 is negative, the right-hand side must also be negative. The square root of a real quantity is always non-negative (assuming we want to avoid imaginary numbers). Thus, we need the negative sign in front so that the sign of both sides of the equation are the same. – 2017-02-28
2 Answers
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To find the sign for the square root in this case you have to consider what the sign of $y$ is at $6$. It is negative so your expression on the right must also be negative so you can't use the principle square root because that's always positive so solution is $y = -\sqrt{2x^2 - 56}$.
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I think instead of finding y and then putting x=6 and y=-4 is not a right method. You have to direct put values in equation you obtain after integrating.
We have,
8 = 36 + c
c = -24