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

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

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y^{''}-y^'=e^{2x}*cos(e^x)

y^{^{''}} - y^{^{'}} = e^{2 x} \cdot cos (e^{x})

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

Solution steps

y^{''} (x) - y^{'} (x) = e^{2 x} \cdot cos (e^{x})

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + 3 + 2 cos^{2} (x)

y^{^{''}} - 3 y^{^{'}} + 10 y = cosh (2 x)

(d^{2} + 4) y = tan^{2} (x)

(d^{3} - 1) \cdot y = e^{x}

y^{^{''}} + 4 y = 0, y^{1} (0) = sin^{2} (x)

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