0 0 votes A curve passes through the point $\left ( x=1,y=0 \right )$ and satisfies the differential equation $\dfrac{\mathrm{dy} }{\mathrm{d} x}=\dfrac{x^{2}+y^{2}}{2y}+\dfrac{y}{x}.$ The equation that describes the curve is $\ln\left (1+\dfrac{y^{2}}{x^{2}}\right)=x-1$ $\dfrac{1}{2}\ln\left (1+\dfrac{y^{2}}{x^{2}}\right)=x-1$ $\ln\left (1+\dfrac{y}{x}\right)=x-1$ $\dfrac{1}{2}\ln\left (1+\dfrac{y}{x}\right)=x-1$ Differential Equations gate2018-ec differential-equations + – gatecse 548 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.
