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

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

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

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

y = (4 - c_{2}) e^{t} + c_{2} e^{\frac{t}{4}}

Solution steps

4 y^{''} (t) - 5 y^{'} (t) + y = 0

Solve linear ODE:

y = (4 - c_{2}) e^{t} + c_{2} e^{\frac{t}{4}}

y = (4 - c_{2}) e^{t} + c_{2} e^{\frac{t}{4}}

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

a^{2} \cdot \frac{d ^{2} y}{d x ^{2}} + a \cdot \frac{d y}{d x} - c \cdot y = 0

y^{^{'''}} - 11 y^{^{''}} + 24 y^{^{'}} + 36 y = 0, y (0) = 0

y^{^{''''}} + 8 y^{^{''}} - 4 y^{^{'''}} - 16 y^{^{'}} + 16 y = 0

10 y^{^{''}} + 30 y^{^{'}} + 20 y = 0

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