What is the solution for y^{''}-3/2 y^'+y=0 ?

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

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y^{''}-3/2 y^'+y=0

y^{^{''}} - \frac{3}{2} y^{^{'}} + y = 0

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

Solution steps

y^{''} (t) - \frac{3}{2} y^{'} (t) + y = 0

Solve linear ODE:

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

77 y^{^{''''}} + 39 y^{^{''}} + 4 y = 0

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

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

3 \frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} - 11 y = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (1) = e^{3}

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