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}

5 3 x^{3} + x

\frac{e ^{\frac{1}{x}}}{x ^{3}}

f (x) = x + 1, (3, 2)

\int C e^{- a t} d t

f (x) = \frac{6}{x ^{4}}

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