What is the solution for y^{''}+7y=0 ?

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

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y^{''}+7y=0

y^{^{''}} + 7 y = 0

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

Solution steps

y^{''} (t) + 7 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (e^{3 x} + 3 x e^{3 x})

\frac{d}{d x} (2 x^{4 + 7 x + 5})

\int \frac{y}{( 1 + y ^{2} )} d y

\int \frac{4 x ^{2} - 3 x + 4}{x ^{3} + 4 x} d x

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- x}

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