What is the solution for y-xy^'=1992(1+x^2y^') ?

The solution for y-xy^'=1992(1+x^2y^') is y=-(c_{1}x)/(1992x+1)+1992

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y-xy^'=1992(1+x^2y^')

y - x y^{^{'}} = 1992 (1 + x^{2} y^{^{'}})

y = - \frac{c _{1} x}{1992 x + 1} + 1992

Solution steps

y - x y^{'} (x) = 1992 (1 + x^{2} y^{'} (x))

Solve separable ODE:

y = - \frac{c _{1} x}{1992 x + 1} + 1992

y = - \frac{c _{1} x}{1992 x + 1} + 1992

sin (x) \frac{d y}{d x} + cos (x) y = 0

(x - 1) y^{^{'}} = 4 - y^{2}

y^{^{'}} + y = 2 y^{\frac{1}{2}}

\frac{d}{d θ} (r (θ)) + r (θ) tan (θ) = 3 tan (θ)

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

What is the solution for y-xy^'=1992(1+x^2y^') ?
The solution for y-xy^'=1992(1+x^2y^') is y=-(c_{1}x)/(1992x+1)+1992