What is the solution for y^{''}+16y=5cos(4t),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+16y=5cos(4t),y(0)=1,y^'(0)=0 is y=(5tsin(4t))/8+cos(4t)

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y^{''}+16y=5cos(4t),y(0)=1,y^'(0)=0

y^{^{''}} + 16 y = 5 cos (4 t), y (0) = 1, y^{^{'}} (0) = 0

y = \frac{5 t sin ( 4 t )}{8} + cos (4 t)

Solution steps

y^{''} (t) + 16 y = 5 cos (4 t)

Solve linear ODE:

y = \frac{5 t sin ( 4 t )}{8} + cos (4 t)

y = \frac{5 t sin ( 4 t )}{8} + cos (4 t)

y^{^{''}} - 9 y = 4, y^{1} (0) = e^{- 3 x}

y^{^{''''}} + 9 y^{^{''}} = 20 e^{- t}

y^{^{''''}} - y = 9 t + cos (t)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 4 e^{2 x}

4 y^{^{''}} - 8 y = 2

What is the solution for y^{''}+16y=5cos(4t),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+16y=5cos(4t),y(0)=1,y^'(0)=0 is y=(5tsin(4t))/8+cos(4t)