What is the solution for y^{''}-5y^'+4y=cos(x) ?

The solution for y^{''}-5y^'+4y=cos(x) is y=c_{1}e^{4x}+c_{2}e^x-5/34 sin(x)+3/34 cos(x)

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y^{''}-5y^'+4y=cos(x)

y^{^{''}} - 5 y^{^{'}} + 4 y = cos (x)

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{5}{34} sin (x) + \frac{3}{34} cos (x)

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = cos (x)

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{5}{34} sin (x) + \frac{3}{34} cos (x)

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{5}{34} sin (x) + \frac{3}{34} cos (x)

y^{^{''}} + 9 y = 2 e^{3 t} + 2 cos (2 t)

y^{^{''}} - y^{^{'}} = 8 e x

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = \frac{e ^{x}}{x}

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

5 y^{^{''}} - 40 y^{^{'}} + 35 y = 26 x

What is the solution for y^{''}-5y^'+4y=cos(x) ?
The solution for y^{''}-5y^'+4y=cos(x) is y=c_{1}e^{4x}+c_{2}e^x-5/34 sin(x)+3/34 cos(x)