What is the solution for y^{''}+5y+4=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+5y+4=0,y(0)=1,y^'(0)=1 is y= 9/5 cos(sqrt(5)t)+1/(sqrt(5))sin(sqrt(5)t)-4/5

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y^{''}+5y+4=0,y(0)=1,y^'(0)=1

y^{^{''}} + 5 y + 4 = 0, y (0) = 1, y^{^{'}} (0) = 1

y = \frac{9}{5} cos (5 t) + \frac{1}{5} sin (5 t) - \frac{4}{5}

Solution steps

y^{''} (t) + 5 y + 4 = 0

Solve linear ODE:

y = \frac{9}{5} cos (5 t) + \frac{1}{5} sin (5 t) - \frac{4}{5}

y = \frac{9}{5} cos (5 t) + \frac{1}{5} sin (5 t) - \frac{4}{5}

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

y^{^{''}} + 4 y^{^{'}} = t^{2} + 2 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 3 y = 8 sin (t)

\frac{d ^{2} y}{d t ^{2}} - y = 10 e^{t}

y^{^{''}} - 0 \cdot y^{^{'}} - 2 y = - 2 t + 4 t^{2}

What is the solution for y^{''}+5y+4=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+5y+4=0,y(0)=1,y^'(0)=1 is y= 9/5 cos(sqrt(5)t)+1/(sqrt(5))sin(sqrt(5)t)-4/5