What is the solution for y^{''}-2y^'+y=e^xsec^2(x) ?

The solution for y^{''}-2y^'+y=e^xsec^2(x) is y=c_{1}e^x+c_{2}xe^x-e^xln(cos(x))

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y^{''}-2y^'+y=e^xsec^2(x)

y^{^{''}} - 2 y^{^{'}} + y = e^{x} sec^{2} (x)

y = c_{1} e^{x} + c_{2} x e^{x} - e^{x} ln (cos (x))

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = e^{x} sec^{2} (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - e^{x} ln (cos (x))

y = c_{1} e^{x} + c_{2} x e^{x} - e^{x} ln (cos (x))

\frac{1}{4} y^{^{''}} + 16 y = 8 sin (8 t)

y^{^{''}} - 9 y = sin (x) + 2 cos (x)

y^{^{''}} - 5 y^{^{'}} + 6 y = (x + 1) e^{x}

y^{^{''}} - 3 y^{^{'}} + 2 y = 10

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 (x + 1)

What is the solution for y^{''}-2y^'+y=e^xsec^2(x) ?
The solution for y^{''}-2y^'+y=e^xsec^2(x) is y=c_{1}e^x+c_{2}xe^x-e^xln(cos(x))