What is the solution for y^{''}+c*y=g(x),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+c*y=g(x),y(0)=0,y^'(0)=0 is y=0*sin(sqrt(c)x)

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y^{''}+c*y=g(x),y(0)=0,y^'(0)=0

y^{^{''}} + c \cdot y = g (x), y (0) = 0, y^{^{'}} (0) = 0

y = 0 \cdot sin (c x)

Solution steps

y^{''} (x) + c \cdot y = g (x)

Solve linear ODE:

y = 0 \cdot sin (c x)

y = 0 \cdot sin (c x)

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{x}, y (0) = 0, y^{^{'}} (0) = a

y^{^{''}} - 36 y = 24 sin (6 t)

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

y^{^{''}} + 7 y^{^{'}} + 11 y = 11 x^{3} + 10 x^{2} - 8 x + 9

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 9} e^{5 t}

What is the solution for y^{''}+c*y=g(x),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+c*y=g(x),y(0)=0,y^'(0)=0 is y=0*sin(sqrt(c)x)