What is the solution for 4y^{''}-8y^'+3y=0 ?

The solution for 4y^{''}-8y^'+3y=0 is y=c_{1}e^{(3t)/2}+c_{2}e^{t/2}

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4y^{''}-8y^'+3y=0

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0

y = c_{1} e^{\frac{3 t}{2}} + c_{2} e^{\frac{t}{2}}

Solution steps

4 y^{''} (t) - 8 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{3 t}{2}} + c_{2} e^{\frac{t}{2}}

y = c_{1} e^{\frac{3 t}{2}} + c_{2} e^{\frac{t}{2}}

12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = 0

(D^{6} - 4 D^{4} + 4 D^{2}) y = 0

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

\frac{d ^{2} y}{d x ^{2}} = 16 y

y^{^{''}} + 2 y^{^{'}} + y = 5 y - y^{^{''}} + 2 y^{^{'}} - y

What is the solution for 4y^{''}-8y^'+3y=0 ?
The solution for 4y^{''}-8y^'+3y=0 is y=c_{1}e^{(3t)/2}+c_{2}e^{t/2}