What is the solution for y^{''}-y=(2-4x)e^{-x}+10cos(2x) ?

The solution for y^{''}-y=(2-4x)e^{-x}+10cos(2x) is y=c_{1}e^x+c_{2}e^{-x}+e^{-x}x^2-2cos(2x)

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y^{''}-y=(2-4x)e^{-x}+10cos(2x)

y^{^{''}} - y = (2 - 4 x) e^{- x} + 10 cos (2 x)

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

Solution steps

y^{''} (x) - y = (2 - 4 x) e^{- x} + 10 cos (2 x)

Solve linear ODE:

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

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

f (x) = 1 - sin (x)

\frac{d}{d x} (- x + \frac{4}{x} + 1)

\frac{t ^{2} - 6 t}{( t - 3 ) ^{2}}

\int sec (t) (3 sec (t) + 5 tan (t)) d t

What is the solution for y^{''}-y=(2-4x)e^{-x}+10cos(2x) ?
The solution for y^{''}-y=(2-4x)e^{-x}+10cos(2x) is y=c_{1}e^x+c_{2}e^{-x}+e^{-x}x^2-2cos(2x)