What is the (dy)/(dx)=(ycos(x))/(7+y^2),y(0)=1 ?

The (dy)/(dx)=(ycos(x))/(7+y^2),y(0)=1 is (cos(x))/8

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(dy)/(dx)=(ycos(x))/(7+y^2),y(0)=1

\frac{d y}{d x} = \frac{y cos ( x )}{7 + y ^{2}}, y (0) = 1

\frac{cos ( x )}{8}

Solution steps

\frac{y cos ( x )}{7 + y ^{2}} = \frac{y cos ( x )}{7 + y ^{2}}

Constant function results in a constant

y (0) = 1

= \frac{1 \cdot cos ( x )}{7 + 1 ^{2}}

Expand

\frac{1 \cdot cos ( x )}{7 + 1 ^{2}} : \frac{cos ( x )}{8}

= \frac{cos ( x )}{8}

3 e^{t} tan (y) d t + (2 - e^{t}) sec^{2} (y) d y = 0

y^{^{'}} = 2 x^{3} y

y^{^{'}} y = t, y (3) = - 3

\frac{d x}{d y} = e^{2 x + 3 y}, x (0) = 1

(3 x - 2) d x + (4 y + b) d y = 0

What is the (dy)/(dx)=(ycos(x))/(7+y^2),y(0)=1 ?
The (dy)/(dx)=(ycos(x))/(7+y^2),y(0)=1 is (cos(x))/8