What is the solution for y^{''}+16y=16,y(0)=1,y^'(0)=4 ?

The solution for y^{''}+16y=16,y(0)=1,y^'(0)=4 is y=sin(4t)+1

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y^{''}+16y=16,y(0)=1,y^'(0)=4

y^{^{''}} + 16 y = 16, y (0) = 1, y^{^{'}} (0) = 4

y = sin (4 t) + 1

Solution steps

y^{''} (t) + 16 y = 16

Solve linear ODE:

y = sin (4 t) + 1

y = sin (4 t) + 1

y^{^{''}} + 5 y^{^{'}} + 4 y = e^{- x} + 2

y^{^{'''}} - 2 y^{^{'}} + 4 y = e x cos (x) + x^{2}

y^{^{''}} + 8 y^{^{'}} + 25 y = 150, y^{^{'}} (0) = 0, y (0) = 0

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 t e^{t}, y (0) = 1, y^{^{'}} (0) = - 4

\frac{d ^{2} y}{d x ^{2}} + 14 \frac{d y}{d x} + 49 y = 4 e^{3 x}

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