What is the solution for (y-t)+(t+2y)y^'=0 ?

The solution for (y-t)+(t+2y)y^'=0 is y=(-t+sqrt(3t^2+c_{1)})/2 ,y=-(t+sqrt(3t^2+c_{1)})/2

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(y-t)+(t+2y)y^'=0

(y - t) + (t + 2 y) y^{^{'}} = 0

y = \frac{- t + 3 t ^{2} + c _{1}}{2}, y = - \frac{t + 3 t ^{2} + c _{1}}{2}

Solution steps

(y - t) + (t + 2 y) y^{'} (t) = 0

Solve exact ODE:

y = \frac{- t + 3 t ^{2} + c _{1}}{2}, y = - \frac{t + 3 t ^{2} + c _{1}}{2}

y = \frac{- t + 3 t ^{2} + c _{1}}{2}, y = - \frac{t + 3 t ^{2} + c _{1}}{2}

\frac{d ^{2} y}{d t ^{2}} + y = 0

\int 6 e^{3 x} d x

x \to 0 lim (\frac{sin ( 3 x )}{3})

x \to 0 lim (\frac{x ^{4} + 2 x}{x})

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

What is the solution for (y-t)+(t+2y)y^'=0 ?
The solution for (y-t)+(t+2y)y^'=0 is y=(-t+sqrt(3t^2+c_{1)})/2 ,y=-(t+sqrt(3t^2+c_{1)})/2