What is the solution for y^{''}-y^'=tan(x) ?

The solution for y^{''}-y^'=tan(x) is y=c_{1}e^x+c_{2}+e^x*integral of (tan(x))/(e^x)+ln(cos(x))

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

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

y = c_{1} e^{x} + c_{2} + e^{x} \cdot \int \frac{tan ( x )}{e ^{x}} d x + ln (cos (x))

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} + e^{x} \cdot \int \frac{tan ( x )}{e ^{x}} d x + ln (cos (x))

y = c_{1} e^{x} + c_{2} + e^{x} \cdot \int \frac{tan ( x )}{e ^{x}} d x + ln (cos (x))

y^{^{''}} + 4 y = \frac{5}{2} cos (2 x)

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = 1 - 4 x^{3}

y^{^{''''}} + 2 y^{^{''}} + y = 7 + cos (2 t)

x^{^{''}} + 6 x^{^{'}} + 34 x = 68, x (0) = 5, x^{^{'}} (0) = 7

y^{^{''}} + 9 y = 13 sec^{2} (3 t)

What is the solution for y^{''}-y^'=tan(x) ?
The solution for y^{''}-y^'=tan(x) is y=c_{1}e^x+c_{2}+e^x*integral of (tan(x))/(e^x)+ln(cos(x))