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

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

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

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

y = c_{1} e^{\frac{7 t}{2}} + c_{2} e^{- \frac{7 t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{7 t}{2}} + c_{2} e^{- \frac{7 t}{2}}

y = c_{1} e^{\frac{7 t}{2}} + c_{2} e^{- \frac{7 t}{2}}

\frac{d ^{2} x}{d t ^{2}} + 10 \frac{d x}{d t} + 25 x = 0

y^{^{''}} + 10 y^{^{'}} + 16 y = 0, y (0) = 9, y^{^{'}} (0) = - 60

y^{^{'''}} - 7 y^{^{''}} + 12 y^{^{'}} = 0

x^{^{''}} + 2 x^{^{'}} - 8 x = 0, x^{^{'}} (0) = 22, x (0) = - 1

\frac{d ^{2} x}{d t ^{2}} - 4 \frac{d x}{d t} + 5 x = 0

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