What is the solution for 3y^{''}+48y=0,y(0)=3,y^'(0)=20 ?

The solution for 3y^{''}+48y=0,y(0)=3,y^'(0)=20 is y=3cos(4t)+5sin(4t)

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3y^{''}+48y=0,y(0)=3,y^'(0)=20

3 y^{^{''}} + 48 y = 0, y (0) = 3, y^{^{'}} (0) = 20

y = 3 cos (4 t) + 5 sin (4 t)

Solution steps

3 y^{''} (t) + 48 y = 0

Solve linear ODE:

y = 3 cos (4 t) + 5 sin (4 t)

y = 3 cos (4 t) + 5 sin (4 t)

5 \frac{d ^{2} x}{d t ^{2}} - 3 x = 0

y^{^{''''}} - 4 y^{^{'''}} - 5 y^{^{''}} + 36 y^{^{'}} - 36 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y^{^{'}} 0 = 10, y^{0} = 2

\frac{d ^{2} f}{d x ^{2}} + a f (x) = 0

1296 y^{^{''}} + 72 y^{^{'}} + y = 0, y (0) = 7, y^{^{'}} (0) = 5

What is the solution for 3y^{''}+48y=0,y(0)=3,y^'(0)=20 ?
The solution for 3y^{''}+48y=0,y(0)=3,y^'(0)=20 is y=3cos(4t)+5sin(4t)