What is the solution for (y^{''})/y =-3 ?

The solution for (y^{''})/y =-3 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)

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(y^{''})/y =-3

\frac{y ^{^{''}}}{y} = - 3

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Solution steps

\frac{y ^{''} ( t )}{y} = - 3

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t)

\int (x^{2} \cdot sin (3 x)) d x

\frac{d y}{d x} + 10 x^{4} y = 50 x^{9} e^{3 x^{5}}

\int 1 + t^{2} d t

\int - 25 sin (- 5 t) d t

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

What is the solution for (y^{''})/y =-3 ?
The solution for (y^{''})/y =-3 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)