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}

x + 2

x^{5} + 2, x + 2

\int sin (m x) d x

\frac{d y}{d x} = (\frac{y - x + 1}{y - x - 6})

f (x) = 1 x - 32 x, at 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