What is the solution for y^{''}+49y=0,y(0)=4,y^{''}(0)=-6 ?

The solution for y^{''}+49y=0,y(0)=4,y^{''}(0)=-6 is y=4cos(7t)+c_{2}sin(7t)

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y^{''}+49y=0,y(0)=4,y^{''}(0)=-6

y^{^{''}} + 49 y = 0, y (0) = 4, y^{^{''}} (0) = - 6

y = 4 cos (7 t) + c_{2} sin (7 t)

Solution steps

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

Solve linear ODE:

y = 4 cos (7 t) + c_{2} sin (7 t)

y = 4 cos (7 t) + c_{2} sin (7 t)

\int_{0}^{1} 4 (1 + x)^{4} d x

f (x) = 2 x + 2

\frac{\partial}{\partial x} (arctan (x^{2} y))

\int 5 x x^{2} + 1 d x

y^{^{''}} - 2 y^{^{'}} - 11 y = 0

What is the solution for y^{''}+49y=0,y(0)=4,y^{''}(0)=-6 ?
The solution for y^{''}+49y=0,y(0)=4,y^{''}(0)=-6 is y=4cos(7t)+c_{2}sin(7t)