Answer:
Mathematically it is given that
[tex]\frac{dy}{dx}=\frac{1}{y}\\\\ydy=dx...........(i)\\Integrating\\\\\int ydy=\int dx\\\\\frac{y^{2}}{2}=x+c[/tex] where 'c' is a constant
equation i is the required differential equation
thus the solution becomes
[tex]\therefore y=\sqrt{2x+c}[/tex]
Now we have to verify that [tex]y=-\sqrt{2x+c}[/tex] is a solution of the differential equation
Thus differentiating it with respect to 'x' we get
[tex]\frac{dy}{dx}=\frac{d(-\sqrt{2x+c})}{dx}\\\\\frac{dy}{dx}=\frac{-1}{2\sqrt{2x+c}}\times 2\\\\\therefore \frac{dy}{dx}=\frac{1}{y}[/tex]
Hence verified
