What is the solution for 16y^{''}-8y^'+y=0,y(0)=-4,y^'(0)=3 ?

The solution for 16y^{''}-8y^'+y=0,y(0)=-4,y^'(0)=3 is y=-4e^{t/4}+4e^{t/4}t

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16y^{''}-8y^'+y=0,y(0)=-4,y^'(0)=3

16 y^{^{''}} - 8 y^{^{'}} + y = 0, y (0) = - 4, y^{^{'}} (0) = 3

y = - 4 e^{\frac{t}{4}} + 4 e^{\frac{t}{4}} t

Solution steps

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

Solve linear ODE:

y = - 4 e^{\frac{t}{4}} + 4 e^{\frac{t}{4}} t

y = - 4 e^{\frac{t}{4}} + 4 e^{\frac{t}{4}} t

y^{^{''}} - y^{^{'}} - 12 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

y^{^{''}} + 11 y^{^{'}} + 30 y = 0

x^{^{''}} + 5 x^{^{''}} + 6 x = 0

y^{^{''''}} - 2 y^{^{'''}} - 7 y^{^{''}} = 0

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} - 20 y = 0

What is the solution for 16y^{''}-8y^'+y=0,y(0)=-4,y^'(0)=3 ?
The solution for 16y^{''}-8y^'+y=0,y(0)=-4,y^'(0)=3 is y=-4e^{t/4}+4e^{t/4}t