What is the solution for 4y^{''}-4y^'+37y=0,y(0)=3,y^'(0)= 3/2 ?

The solution for 4y^{''}-4y^'+37y=0,y(0)=3,y^'(0)= 3/2 is y=3e^{t/2}cos(3t)

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4y^{''}-4y^'+37y=0,y(0)=3,y^'(0)= 3/2

4 y^{^{''}} - 4 y^{^{'}} + 37 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{3}{2}

y = 3 e^{\frac{t}{2}} cos (3 t)

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 37 y = 0

Solve linear ODE:

y = 3 e^{\frac{t}{2}} cos (3 t)

y = 3 e^{\frac{t}{2}} cos (3 t)

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

y^{^{''}} = 8 y^{^{'}} - 15 y, y (0) = 0, y^{^{'}} (0) = - 2

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

0.1 y^{^{''}} + 0.4 y^{^{'}} + 2 y = 0

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

What is the solution for 4y^{''}-4y^'+37y=0,y(0)=3,y^'(0)= 3/2 ?
The solution for 4y^{''}-4y^'+37y=0,y(0)=3,y^'(0)= 3/2 is y=3e^{t/2}cos(3t)