What is the solution for y^{''}-6y+9y=0 ?

The solution for y^{''}-6y+9y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)

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y^{''}-6y+9y=0

y^{^{''}} - 6 y + 9 y = 0

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

Solution steps

y^{''} (t) - 6 y + 9 y = 0

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)

\frac{d y}{d x} = \frac{( 2 x ^{2} + 3 ) ( 3 x + 2 )}{7 x - 3}

(x + 1) \frac{d y}{d x} = x + 6

\frac{d}{d x} (co tan (2 x))

\int \frac{1}{7 + 9 x ^{2}} d x

(2 x - 1) d x + (3 x + 7) d y = 0

What is the solution for y^{''}-6y+9y=0 ?
The solution for y^{''}-6y+9y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)