What is the solution for (dy)/(dx)=(x+y^2)/(2y),y(0)=1 ?

The solution for (dy)/(dx)=(x+y^2)/(2y),y(0)=1 is y=sqrt(-x-1+2e^x)

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(dy)/(dx)=(x+y^2)/(2y),y(0)=1

\frac{d y}{d x} = \frac{x + y ^{2}}{2 y}, y (0) = 1

y = - x - 1 + 2 e^{x}

Solution steps

\frac{d y}{d x} = \frac{x + y ^{2}}{2 y}

Solve Bernoulli ODE:

y = - x - 1 + 2 e^{x}

y = - x - 1 + 2 e^{x}

\int \frac{x ^{5}}{x ^{6} - 7} d x

θ \to 0 lim (\frac{1}{11})

l a pl a ce t r an s f or m sin^{2} (x)

\frac{d}{d x} ((x + 2 x) e^{x})

\frac{d}{d x} (\frac{sin ( x )}{2 x})

What is the solution for (dy)/(dx)=(x+y^2)/(2y),y(0)=1 ?
The solution for (dy)/(dx)=(x+y^2)/(2y),y(0)=1 is y=sqrt(-x-1+2e^x)