What is the solution for y^{''}+3y=5,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+3y=5,y(0)=1,y^'(0)=1 is y=-2/3 cos(sqrt(3)t)+1/(sqrt(3))sin(sqrt(3)t)+5/3

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y^{''}+3y=5,y(0)=1,y^'(0)=1

y^{^{''}} + 3 y = 5, y (0) = 1, y^{^{'}} (0) = 1

y = - \frac{2}{3} cos (3 t) + \frac{1}{3} sin (3 t) + \frac{5}{3}

Solution steps

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

Solve linear ODE:

y = - \frac{2}{3} cos (3 t) + \frac{1}{3} sin (3 t) + \frac{5}{3}

y = - \frac{2}{3} cos (3 t) + \frac{1}{3} sin (3 t) + \frac{5}{3}

y^{^{''}} + 4 y = sin (x) sin (2 x)

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

y^{^{''}} - 2 y^{^{'}} + 10 y = 40, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 \cdot y = - 3

u^{^{''}} - a \cdot H (x) = x

What is the solution for y^{''}+3y=5,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+3y=5,y(0)=1,y^'(0)=1 is y=-2/3 cos(sqrt(3)t)+1/(sqrt(3))sin(sqrt(3)t)+5/3