What is the solution for y^{''}+4y=3,y^'(0)=0,y(0)=0 ?

The solution for y^{''}+4y=3,y^'(0)=0,y(0)=0 is y=-3/4 cos(2t)+3/4

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y^{''}+4y=3,y^'(0)=0,y(0)=0

y^{^{''}} + 4 y = 3, y^{^{'}} (0) = 0, y (0) = 0

y = - \frac{3}{4} cos (2 t) + \frac{3}{4}

Solution steps

y^{''} (t) + 4 y = 3

Solve linear ODE:

y = - \frac{3}{4} cos (2 t) + \frac{3}{4}

y = - \frac{3}{4} cos (2 t) + \frac{3}{4}

y^{^{''}} - \frac{7}{4} y^{^{'}} + \frac{3}{4} y = t e^{2 t} sin (- 2 t)

x^{^{''}} - x = - 4 e^{- t}

\frac{d ^{2} y}{d x ^{2}} + y = e^{x}

y^{^{''}} + y^{^{'}} + 6 y = cos (2 x)

0.1 x^{^{''}} + 10 x = 100 sin (c t)

What is the solution for y^{''}+4y=3,y^'(0)=0,y(0)=0 ?
The solution for y^{''}+4y=3,y^'(0)=0,y(0)=0 is y=-3/4 cos(2t)+3/4