What is the solution for 4y^{''}+6y^'+7y=0 ?

The solution for 4y^{''}+6y^'+7y=0 is y=e^{-(3t)/4}(c_{1}cos((sqrt(19)t)/4)+c_{2}sin((sqrt(19)t)/4))

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4y^{''}+6y^'+7y=0

4 y^{^{''}} + 6 y^{^{'}} + 7 y = 0

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{19 t}{4}) + c_{2} sin (\frac{19 t}{4}))

Solution steps

4 y^{''} (t) + 6 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{19 t}{4}) + c_{2} sin (\frac{19 t}{4}))

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{19 t}{4}) + c_{2} sin (\frac{19 t}{4}))

\frac{d}{d x} (sin (e^{2 x y}) + e^{s i n (2 x y)})

t an g e n t y = 4 x + 33, (4, 7)

\int (2 x)^{\frac{1}{2}} d x

\int x^{2} 1 + (2 x)^{2} d x

\int \frac{1}{cot ( 3 x )} d x

What is the solution for 4y^{''}+6y^'+7y=0 ?
The solution for 4y^{''}+6y^'+7y=0 is y=e^{-(3t)/4}(c_{1}cos((sqrt(19)t)/4)+c_{2}sin((sqrt(19)t)/4))