What is the solution for 2y^{''}-4y^'+10y=2,xy(0)=0,y^'(0)=0 ?

The solution for 2y^{''}-4y^'+10y=2,xy(0)=0,y^'(0)=0 is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+1/5

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2y^{''}-4y^'+10y=2,xy(0)=0,y^'(0)=0

2 y^{^{''}} - 4 y^{^{'}} + 10 y = 2, x y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = t - 2, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} + 5 y = e^{2 t} \cdot 3

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

y^{^{''}} + 9 y^{^{'}} = e^{x}, y (0) = 0, y^{^{'}} (0) = 0

3 y^{^{''}} - 5 y^{^{'}} + y = x^{2} + 1 + sin (2 x)

What is the solution for 2y^{''}-4y^'+10y=2,xy(0)=0,y^'(0)=0 ?
The solution for 2y^{''}-4y^'+10y=2,xy(0)=0,y^'(0)=0 is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+1/5