What is the solution for y^{''}+y=5t^2-5,y(0)=0,y^'(0)=-6 ?

The solution for y^{''}+y=5t^2-5,y(0)=0,y^'(0)=-6 is y=15cos(t)-6sin(t)+5t^2-15

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y^{''}+y=5t^2-5,y(0)=0,y^'(0)=-6

y^{^{''}} + y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 6

y = 15 cos (t) - 6 sin (t) + 5 t^{2} - 15

Solution steps

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

Solve linear ODE:

y = 15 cos (t) - 6 sin (t) + 5 t^{2} - 15

y = 15 cos (t) - 6 sin (t) + 5 t^{2} - 15

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 9

y^{^{''}} - 8 y^{^{'}} + 16 y = 24 x + 2

\frac{d ^{2} y}{d x ^{2}} + y = - 16 cos (3 x)

y^{^{''}} - 4 y^{^{'}} + 3 y = \frac{1}{1 + e ^{- x}}

x^{^{''}} + 8 x^{^{'}} + 16 x = 2 sin (4 t)

What is the solution for y^{''}+y=5t^2-5,y(0)=0,y^'(0)=-6 ?
The solution for y^{''}+y=5t^2-5,y(0)=0,y^'(0)=-6 is y=15cos(t)-6sin(t)+5t^2-15