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

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

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

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

y = c_{1} cos (t) - 3 sin (t) + 1

Solution steps

y^{''} (t) + y = 1

Solve linear ODE:

y = c_{1} cos (t) - 3 sin (t) + 1

y = c_{1} cos (t) - 3 sin (t) + 1

y^{^{''}} + 4 y = t cos (2 t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 5 e^{4} x, y (0) = 1, y^{^{'}} (0) = 0

9 y^{^{''}} + 4 y = 3 x + 12 e^{- x}

y^{^{''''}} + 8 y^{^{''}} + 16 y = - sin (x)

x^{^{''}} + 6 x^{^{'}} + 23 x = 12 cos (4 t)

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