What is expand y'(x)-(y^2+1)sin(x) ?

The solution to expand y'(x)-(y^2+1)sin(x) is y'(x)-sin(x)y^2-sin(x)

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expand y'(x)-(y^2+1)sin(x)

expand

y^{'} (x) - (y^{2} + 1) sin (x)

y' (x) - sin (x) y^{2} - sin (x)

Solution steps

y^{'} (x) - (y^{2} + 1) sin (x)

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

Expand

- sin (x) (y^{2} + 1) : - sin (x) y^{2} - sin (x)

= y^{'} (x) - sin (x) y^{2} - sin (x)

\frac{( p \cdot ( l - a ) )}{l}

(2 - 0.1 t^{2}) d t

y (0.8 x + 0.2 y = 0)

\frac{( x + a ) ^{2} - a}{a}

\frac{( - x + 15 ) ( - x + 10 )}{2}

What is expand y'(x)-(y^2+1)sin(x) ?
The solution to expand y'(x)-(y^2+1)sin(x) is y'(x)-sin(x)y^2-sin(x)