What is the (dy)/(dx)=(3y)/(2y^2+1),y=1,x=3 ?

The (dy)/(dx)=(3y)/(2y^2+1),y=1,x=3 is 1

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(dy)/(dx)=(3y)/(2y^2+1),y=1,x=3

\frac{d y}{d x} = \frac{3 y}{2 y ^{2} + 1}, y = 1, x = 3

1

Solution steps

Constant function results in a constant

y = 1

= \frac{3 \cdot 1}{2 \cdot 1 ^{2} + 1}

\frac{3 \cdot 1}{2 \cdot 1 ^{2} + 1} = 1

= 1

(t^{2} - 16 t + 15) \frac{d y}{d t} = y

x^{2} (x d y + y d x) = 1 x d y - 1 y d x

y^{^{'}} = x y^{3} (1 + x^{2})^{\frac{1}{2}}

\frac{d y}{d x} = \frac{y ln ( y )}{x}

x y^{3} d x + e^{- x} d y = 0

What is the (dy)/(dx)=(3y)/(2y^2+1),y=1,x=3 ?
The (dy)/(dx)=(3y)/(2y^2+1),y=1,x=3 is 1