What is the solution for (D^2-6D+13)y=5^x ?

The solution for (D^2-6D+13)y=5^x is

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(D^2-6D+13)y=5^x

(D^{2} - 6 D + 13) y = 5^{x}

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

Solution steps

(D^{2} - 6 D + 13) y = 5^{x}

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 10 cos (5 x)

y^{^{''}} + y = 2 x + 4

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = sin (x)

y^{^{''}} + 16 y = - 16 sec (4 t)

\frac{d ^{2} y}{d x ^{2}} - 10 \cdot \frac{d y}{d x} + 25 \cdot y = 30 \cdot x + 3

What is the solution for (D^2-6D+13)y=5^x ?
The solution for (D^2-6D+13)y=5^x is