What is the solution for y^{''}+4y-3y^'=0 ?

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

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y^{''}+4y-3y^'=0

y^{^{''}} + 4 y - 3 y^{^{'}} = 0

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

y^{''} (t) + 4 y - 3 y^{'} (t) = 0

Solve linear ODE:

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

- 100 y + 65 y^{^{'}} - 14 y^{^{''}} + y^{^{'''}} = 0

x^{^{''}} + 4 x^{^{'}} = - 2 x^{^{'}} - 8 x + 3 x

12 y^{^{''}} - 5 y - 2 y = 0

y^{^{''}} + 2 y = 0, y (0) = sin (k t)

\frac{d ^{2} y}{d t ^{2}} = 5 \frac{d y}{d t} - 4 y

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