What is the solution for y^{''}+4y=sin(t),y(0)=1,y^'(0)=-1 ?

The solution for y^{''}+4y=sin(t),y(0)=1,y^'(0)=-1 is y=cos(2t)-2/3 sin(2t)+1/3 sin(t)

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y^{''}+4y=sin(t),y(0)=1,y^'(0)=-1

y^{^{''}} + 4 y = sin (t), y (0) = 1, y^{^{'}} (0) = - 1

y = cos (2 t) - \frac{2}{3} sin (2 t) + \frac{1}{3} sin (t)

Solution steps

y^{''} (t) + 4 y = sin (t)

Solve linear ODE:

y = cos (2 t) - \frac{2}{3} sin (2 t) + \frac{1}{3} sin (t)

y = cos (2 t) - \frac{2}{3} sin (2 t) + \frac{1}{3} sin (t)

y^{^{''}} - 4 y = 5 e^{2 x} + 6, y^{^{'}} (0) = - 1, y (0) = 0

y^{^{'''}} + y^{^{''}} - 2 y = x e^{x} + 15

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + \frac{1}{4} y = x e^{\frac{x}{2}}

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

3 y^{^{''}} - 2 y^{^{'}} - y = x^{2}

What is the solution for y^{''}+4y=sin(t),y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+4y=sin(t),y(0)=1,y^'(0)=-1 is y=cos(2t)-2/3 sin(2t)+1/3 sin(t)