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)-6/7 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) - \frac{6}{7} sin (7 t)

Solution steps

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

Solve linear ODE:

y = 4 cos (7 t) - \frac{6}{7} sin (7 t)

y = 4 cos (7 t) - \frac{6}{7} sin (7 t)

y^{^{'}} - y = 3 x - 3

f (x) = 1 - \frac{1}{x + 2}

- \frac{3}{2} x^{2} + 3 x + 3

\int \frac{1}{r} d r

\frac{d}{d x} (\frac{x ^{2}}{9 + 2 x})

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)-6/7 sin(7t)