What is the solution for y^'=(4t+y)/(4y-t) ?

The solution for y^'=(4t+y)/(4y-t) is y=(t+sqrt(17t^2+c_{1)})/4 ,y=(t-sqrt(17t^2+c_{1)})/4

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y^'=(4t+y)/(4y-t)

y^{^{'}} = \frac{4 t + y}{4 y - t}

y = \frac{t + 17 t ^{2} + c _{1}}{4}, y = \frac{t - 17 t ^{2} + c _{1}}{4}

Solution steps

y^{'} (t) = \frac{4 t + y}{4 y - t}

Solve exact ODE:

y = \frac{t + 17 t ^{2} + c _{1}}{4}, y = \frac{t - 17 t ^{2} + c _{1}}{4}

y = \frac{t + 17 t ^{2} + c _{1}}{4}, y = \frac{t - 17 t ^{2} + c _{1}}{4}

\frac{\partial}{\partial y} (sin (x^{2} + y^{2}))

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

\int \frac{1 + cos ( x )}{1 + sin ( x )} d x

\frac{\partial}{\partial x} (5 x^{4} + 2 x y + 1)

e^{2 t} cos (3 t)

What is the solution for y^'=(4t+y)/(4y-t) ?
The solution for y^'=(4t+y)/(4y-t) is y=(t+sqrt(17t^2+c_{1)})/4 ,y=(t-sqrt(17t^2+c_{1)})/4