What is the solution for y^{''}-4y^'-45y=(7+2x) ?

The solution for y^{''}-4y^'-45y=(7+2x) is y=c_{1}e^{9x}+c_{2}e^{-5x}-(2x)/(45)-307/2025

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y^{''}-4y^'-45y=(7+2x)

y^{^{''}} - 4 y^{^{'}} - 45 y = (7 + 2 x)

y = c_{1} e^{9 x} + c_{2} e^{- 5 x} - \frac{2 x}{45} - \frac{307}{2025}

Solution steps

y^{''} (x) - 4 y^{'} (x) - 45 y = (7 + 2 x)

Solve linear ODE:

y = c_{1} e^{9 x} + c_{2} e^{- 5 x} - \frac{2 x}{45} - \frac{307}{2025}

y = c_{1} e^{9 x} + c_{2} e^{- 5 x} - \frac{2 x}{45} - \frac{307}{2025}

y^{^{''}} + 81 y = 2 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - y = x, y (0) = 0, y (1) = 0

y^{^{''}} + 2 y^{^{'}} + 2 y = x + 1

y^{^{''}} - 2 y = sin (t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 34 cos (2 x)

What is the solution for y^{''}-4y^'-45y=(7+2x) ?
The solution for y^{''}-4y^'-45y=(7+2x) is y=c_{1}e^{9x}+c_{2}e^{-5x}-(2x)/(45)-307/2025