What is the solution for y^{''}+49y=0,y(0)=5,y^'(0)=2 ?

The solution for y^{''}+49y=0,y(0)=5,y^'(0)=2 is y=5cos(7t)+2/7 sin(7t)

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y^{''}+49y=0,y(0)=5,y^'(0)=2

y^{^{''}} + 49 y = 0, y (0) = 5, y^{^{'}} (0) = 2

y = 5 cos (7 t) + \frac{2}{7} sin (7 t)

Solution steps

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

Solve linear ODE:

y = 5 cos (7 t) + \frac{2}{7} sin (7 t)

y = 5 cos (7 t) + \frac{2}{7} sin (7 t)

y^{^{''}} + a \cdot y^{^{'}} = 0

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 15 y = 0

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

20 y^{^{''}} + 9 y^{^{'}} + y = 0

What is the solution for y^{''}+49y=0,y(0)=5,y^'(0)=2 ?
The solution for y^{''}+49y=0,y(0)=5,y^'(0)=2 is y=5cos(7t)+2/7 sin(7t)