Question

ydy=4x(y^2+1)^1/2 dx y(0)=1

Answer 1

$ydy = 4x{\left( {{y^2} + 1} \right)^{\frac{1}{2}}}dx,\,\,\,y(0) = 1$

$\frac{{ydy}}{{{{\left( {{y^2} + 1} \right)}^{\frac{1}{2}}}}} = 4xdx$

$\frac{1}{2}\frac{{d\left( {{y^2} + 1} \right)}}{{{{\left( {{y^2} + 1} \right)}^{\frac{1}{2}}}}} = 4xdx$

$\sqrt {{y^2} + 1} = 2{x^2} + C$

$\sqrt {{y^2} + 1} - 2{x^2} = C$

$y(0) = 1 \Rightarrow \sqrt {{1^2} + 1} - 0 = C \Rightarrow C = \sqrt 2$

Answer: $\sqrt {{y^2} + 1} - 2{x^2} = \sqrt 2$