What is the solution for tan(t)y^'-y=1,y(pi/6)=3 ?

The solution for tan(t)y^'-y=1,y(pi/6)=3 is y=8sin(t)-1

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tan(t)y^'-y=1,y(pi/6)=3

tan (t) y^{^{'}} - y = 1, y (\frac{π}{6}) = 3

y = 8 sin (t) - 1

Solution steps

tan (t) y^{'} (t) - y = 1

Solve separable ODE:

y = 8 sin (t) - 1

y = 8 sin (t) - 1

\frac{x}{x ^{2} - x + 1}

y^{^{''}} - 2 y^{^{'}} + y = e^{t}

\frac{d}{d x} (\frac{7 x}{x + 4})

\int \frac{x ^{3}}{x - 2} d x

\int \frac{8}{x ^{2} + 2} d x

What is the solution for tan(t)y^'-y=1,y(pi/6)=3 ?
The solution for tan(t)y^'-y=1,y(pi/6)=3 is y=8sin(t)-1