What is the solution for y^{''}+y=-3,y(0)=0,y^'(0)=1 ?

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

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y^{''}+y=-3,y(0)=0,y^'(0)=1

y^{^{''}} + y = - 3, y (0) = 0, y^{^{'}} (0) = 1

y = 3 cos (t) + sin (t) - 3

Solution steps

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

Solve linear ODE:

y = 3 cos (t) + sin (t) - 3

y = 3 cos (t) + sin (t) - 3

4 y^{^{''}} - y = \frac{e x}{2 + 5}

y^{^{''}} + 2 y^{^{'}} + 2 y = 4 e^{- x} cos (x)

\frac{d ^{2} x}{d t ^{2}} + 6 \frac{d x}{d t} - 5 x = e^{- t}

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

y^{^{''}} - a y = - c

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