What is the solution for y^'=ysec(x) ?

The solution for y^'=ysec(x) is y=e^{c_{1}}tan(x)+e^{c_{1}}sec(x)

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y^'=ysec(x)

y^{^{'}} = y sec (x)

y = e^{c_{1}} tan (x) + e^{c_{1}} sec (x)

Solution steps

y^{'} (x) = y sec (x)

Solve separable ODE:

y = e^{c_{1}} tan (x) + e^{c_{1}} sec (x)

y = e^{c_{1}} tan (x) + e^{c_{1}} sec (x)

\frac{d y}{d x} = y + \frac{y}{x ln ( x )}

\frac{d y}{d t} + 20 y = 24, y = \frac{6}{5} - \frac{6}{5} e^{- 20 t}

d y = 1 - 4 y^{2} d x

y^{^{'}} = cos (x) tan (y)

\frac{d y}{d t} = \frac{t e ^{t}}{y 1 + y ^{2}}

What is the solution for y^'=ysec(x) ?
The solution for y^'=ysec(x) is y=e^{c_{1}}tan(x)+e^{c_{1}}sec(x)