What is the solution for y^{''}+5y=4,y(0)=0,y^'(0)=2 ?

The solution for y^{''}+5y=4,y(0)=0,y^'(0)=2 is y=-4/5 cos(sqrt(5)t)+2/(sqrt(5))sin(sqrt(5)t)+4/5

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y^{''}+5y=4,y(0)=0,y^'(0)=2

y^{^{''}} + 5 y = 4, y (0) = 0, y^{^{'}} (0) = 2

y = - \frac{4}{5} cos (5 t) + \frac{2}{5} sin (5 t) + \frac{4}{5}

Solution steps

y^{''} (t) + 5 y = 4

Solve linear ODE:

y = - \frac{4}{5} cos (5 t) + \frac{2}{5} sin (5 t) + \frac{4}{5}

y = - \frac{4}{5} cos (5 t) + \frac{2}{5} sin (5 t) + \frac{4}{5}

y^{^{''}} - y^{^{'}} - 20 y = - 56 t + 40 t^{2}

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

y^{^{''}} + y^{^{'}} + y = e^{x}

y^{^{''}} + y^{^{'}} = e^{t}

y^{^{''}} + 5 y = 3 cos (2 t)

What is the solution for y^{''}+5y=4,y(0)=0,y^'(0)=2 ?
The solution for y^{''}+5y=4,y(0)=0,y^'(0)=2 is y=-4/5 cos(sqrt(5)t)+2/(sqrt(5))sin(sqrt(5)t)+4/5