What is the solution for y^{''}+16y=50e^3t,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+16y=50e^3t,y(0)=0,y^'(0)=0 is y=(25e^3t)/8-(25e^3)/(32)sin(4t)

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y^{''}+16y=50e^3t,y(0)=0,y^'(0)=0

y^{^{''}} + 16 y = 50 e^{3} t, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{25 e ^{3} t}{8} - \frac{25 e ^{3}}{32} sin (4 t)

Solution steps

y^{''} (t) + 16 y = 50 e^{3} t

Solve linear ODE:

y = \frac{25 e ^{3} t}{8} - \frac{25 e ^{3}}{32} sin (4 t)

y = \frac{25 e ^{3} t}{8} - \frac{25 e ^{3}}{32} sin (4 t)

4 y^{^{''}} - 8 y^{^{'}} + 5 y = e^{t} tan (\frac{t}{2})

(D^{2} + 1) y = 3 sin (2 t) - 2 cos (2 t)

y^{^{''}} + y^{^{'}} = \frac{1}{cos ( x )}

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

y^{^{''}} + y = x + e^{x}

What is the solution for y^{''}+16y=50e^3t,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+16y=50e^3t,y(0)=0,y^'(0)=0 is y=(25e^3t)/8-(25e^3)/(32)sin(4t)