What is the solution for y^{''}-6y^'-3+cos(x)=0 ?

The solution for y^{''}-6y^'-3+cos(x)=0 is y=c_{1}e^{6x}+c_{2}+6/37 sin(x)+1/37 cos(x)-x/2

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y^{''}-6y^'-3+cos(x)=0

y^{^{''}} - 6 y^{^{'}} - 3 + cos (x) = 0

y = c_{1} e^{6 x} + c_{2} + \frac{6}{37} sin (x) + \frac{1}{37} cos (x) - \frac{x}{2}

Solution steps

y^{''} (x) - 6 y^{'} (x) - 3 + cos (x) = 0

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} + \frac{6}{37} sin (x) + \frac{1}{37} cos (x) - \frac{x}{2}

y = c_{1} e^{6 x} + c_{2} + \frac{6}{37} sin (x) + \frac{1}{37} cos (x) - \frac{x}{2}

\frac{d ^{2} y}{d x ^{2}} + y - 1 = 0, y (1) = 0

y^{^{''}} - 4 y = (x^{2} - 5) sin (2 x)

D (x) = - 12 x + 850

y^{^{''}} + b^{2} y = cos (2 t), y (0) = 1

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 t

What is the solution for y^{''}-6y^'-3+cos(x)=0 ?
The solution for y^{''}-6y^'-3+cos(x)=0 is y=c_{1}e^{6x}+c_{2}+6/37 sin(x)+1/37 cos(x)-x/2