What is the solution for y^'=tan(t)y-5t,y(0)=3 ?

The solution for y^'=tan(t)y-5t,y(0)=3 is y=-(5(tsin(t)+cos(t)))/(cos(t))+8/(cos(t))

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y^'=tan(t)y-5t,y(0)=3

y^{^{'}} = tan (t) y - 5 t, y (0) = 3

y = - \frac{5 ( t sin ( t ) + cos ( t ) )}{cos ( t )} + \frac{8}{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{8}{cos ( t )}

y = - \frac{5 ( t sin ( t ) + cos ( t ) )}{cos ( t )} + \frac{8}{cos ( t )}

\frac{d y}{d x} = \frac{e ^{7 x}}{4 y ^{3}}

\frac{\partial}{\partial x} (x^{\frac{1}{3}} \cdot y^{\frac{2}{3}})

y = (ln (x))^{4}

\int \frac{x ^{3}}{( 49 - 4 x ^{2} ) ^{\frac{3}{2}}} d x

What is the solution for y^'=tan(t)y-5t,y(0)=3 ?
The solution for y^'=tan(t)y-5t,y(0)=3 is y=-(5(tsin(t)+cos(t)))/(cos(t))+8/(cos(t))