What is the solution for y^{''}+16y=3sin(4t),y(0)=0,y^'(0)=0 ?

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

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y^{''}+16y=3sin(4t),y(0)=0,y^'(0)=0

y^{^{''}} + 16 y = 3 sin (4 t), y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{3 t cos ( 4 t )}{8} + \frac{3}{32} sin (4 t)

Solution steps

y^{''} (t) + 16 y = 3 sin (4 t)

Solve linear ODE:

y = - \frac{3 t cos ( 4 t )}{8} + \frac{3}{32} sin (4 t)

y = - \frac{3 t cos ( 4 t )}{8} + \frac{3}{32} sin (4 t)

y = x^{3} - 9 x, (- 3, 0)

\int_{0}^{1} 7 e^{x^{2}} d x

\frac{d}{d x} (\frac{2 x ^{2} - 1}{x 1 + x ^{2}})

\int \frac{x ^{2}}{x - 4} d x

5

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