What is the solution for y^{''}+y=3cos(t),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=3cos(t),y(0)=0,y^'(0)=0 is y=(3tsin(t))/2

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y^{''}+y=3cos(t),y(0)=0,y^'(0)=0

y^{^{''}} + y = 3 cos (t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{3 t sin ( t )}{2}

Solution steps

y^{''} (t) + y = 3 cos (t)

Solve linear ODE:

y = \frac{3 t sin ( t )}{2}

y = \frac{3 t sin ( t )}{2}

y^{^{''}} + 6 y^{^{'}} + 5 y = e^{- 3 x}

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

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 54 + 18

d, d (x) = e^{2 x^{3} + 4}

x^{^{''}} + 4 x = sin (3 t)

What is the solution for y^{''}+y=3cos(t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=3cos(t),y(0)=0,y^'(0)=0 is y=(3tsin(t))/2