What is the solution for y^{''}+49y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+49y=0,y(0)=2,y^'(0)=1 is y=2cos(7t)+1/7 sin(7t)

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y^{''}+49y=0,y(0)=2,y^'(0)=1

y^{^{''}} + 49 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = 2 cos (7 t) + \frac{1}{7} sin (7 t)

Solution steps

y^{''} (t) + 49 y = 0

Solve linear ODE:

y = 2 cos (7 t) + \frac{1}{7} sin (7 t)

y = 2 cos (7 t) + \frac{1}{7} sin (7 t)

y^{^{''}} - 4 y + 3 y = 0, y (0) = - 1, y^{^{'}} (0) = - 5

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = 8

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 6, y^{^{'}} (0) = - 11

y^{^{'''}} + y^{^{''}} - 10 y^{^{'}} + 8 y = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y^{^{'}} = 3, y^{0} (0) = 0

What is the solution for y^{''}+49y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+49y=0,y(0)=2,y^'(0)=1 is y=2cos(7t)+1/7 sin(7t)