What is the solution for 4y^{''}-19y^'+12y=0,y(0)=5,y^'(0)=13.5 ?

The solution for 4y^{''}-19y^'+12y=0,y(0)=5,y^'(0)=13.5 is y=3e^{4t}+2e^{(3t)/4}

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4y^{''}-19y^'+12y=0,y(0)=5,y^'(0)=13.5

4 y^{^{''}} - 19 y^{^{'}} + 12 y = 0, y (0) = 5, y^{^{'}} (0) = 13.5

y = 3 e^{4 t} + 2 e^{\frac{3 t}{4}}

Solution steps

4 y^{''} (t) - 19 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = 3 e^{4 t} + 2 e^{\frac{3 t}{4}}

y = 3 e^{4 t} + 2 e^{\frac{3 t}{4}}

What is the solution for 4y^{''}-19y^'+12y=0,y(0)=5,y^'(0)=13.5 ?
The solution for 4y^{''}-19y^'+12y=0,y(0)=5,y^'(0)=13.5 is y=3e^{4t}+2e^{(3t)/4}