What is the solution for y^{''}+2y^'+10y=50x,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+2y^'+10y=50x,y(0)=1,y^'(0)=0 is y=e^{-x}(2cos(3x)-sin(3x))+5x-1

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y^{''}+2y^'+10y=50x,y(0)=1,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + 10 y = 50 x, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 10 y = 50 x

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} = \frac{10 sin ( 7 t )}{2}

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

y^{^{''}} - 4 y^{^{'}} + 3 y = \frac{e ^{t}}{1 + e ^{t}}

\frac{d ^{2} y}{d t ^{2}} + 9 y = 12 cos (3 t)

y^{^{''}} + y = 5 t^{2} - 9, y (0) = 0, y^{^{'}} (0) = - 2

What is the solution for y^{''}+2y^'+10y=50x,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+2y^'+10y=50x,y(0)=1,y^'(0)=0 is y=e^{-x}(2cos(3x)-sin(3x))+5x-1