What is the solution for y^{''}+9y=10,e^{-t}y(0)y^'(0)=0 ?

The solution for y^{''}+9y=10,e^{-t}y(0)y^'(0)=0 is y=c_{1}cos(3t)+c_{2}sin(3t)+10/9

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y^{''}+9y=10,e^{-t}y(0)y^'(0)=0

y^{^{''}} + 9 y = 10, e^{- t} y (0) y^{^{'}} (0) = 0

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{10}{9}

Solution steps

y^{''} (t) + 9 y = 10

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{10}{9}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{10}{9}

y^{^{''}} - 4 y^{^{'}} + 3 y = x cos (x)

y^{^{''}} + y = \frac{cos ( x ) + 1}{sin ( x )}

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

y^{^{''}} + 2 y^{^{'}} - 3 y = sin (t)

(d^{2} - 2 d + 2) y = x^{2} e^{3} x

What is the solution for y^{''}+9y=10,e^{-t}y(0)y^'(0)=0 ?
The solution for y^{''}+9y=10,e^{-t}y(0)y^'(0)=0 is y=c_{1}cos(3t)+c_{2}sin(3t)+10/9