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

The solution for y^{''}-2y^'+y=e^xsec^2(x)tan(x) is y=c_{1}e^x+c_{2}xe^x+(-e^xx+e^xtan(x))/2

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{- e ^{x} x + e ^{x} tan ( x )}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{- e ^{x} x + e ^{x} tan ( x )}{2}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{- e ^{x} x + e ^{x} tan ( x )}{2}

y^{^{''}} - y = 2 t^{2} - t + 3 e^{- 3 t}

q^{^{''}} + 0.1 q^{^{'}} + q = v sin (0.5 t), q (0) = 0

y^{^{''}} - 2 y^{^{'}} = x + e^{x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 30 e^{2 t}

y^{^{''}} + 25 y = 14 cos (5 x) + 21 sin (5 x)

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