What is the solution for y^'=tan(t)y-5t ?

The solution for y^'=tan(t)y-5t is y=-(5(tsin(t)+cos(t)))/(cos(t))+(c_{1})/(cos(t))

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y^'=tan(t)y-5t

y^{^{'}} = tan (t) y - 5 t

y = - \frac{5 ( t sin ( t ) + cos ( t ) )}{cos ( t )} + \frac{c _{1}}{cos ( t )}

Solution steps

y^{'} (t) = tan (t) y - 5 t

Solve linear ODE:

y = - \frac{5 ( t sin ( t ) + cos ( t ) )}{cos ( t )} + \frac{c _{1}}{cos ( t )}

y = - \frac{5 ( t sin ( t ) + cos ( t ) )}{cos ( t )} + \frac{c _{1}}{cos ( t )}

2916 x, 108 x^{2}

\frac{d}{d x} ((4 x)^{4 x})

\frac{\partial}{\partial x} (sin (5 x) cos (5 y))

\frac{d}{d x} (\frac{x ^{3}}{cos ( x )})

\frac{e ^{(- 2 s)}}{s}

What is the solution for y^'=tan(t)y-5t ?
The solution for y^'=tan(t)y-5t is y=-(5(tsin(t)+cos(t)))/(cos(t))+(c_{1})/(cos(t))