What is the solution for 4y^{''}-5y^'+y=0,y(0)=4,y^'(0)= 1/4 ?

The solution for 4y^{''}-5y^'+y=0,y(0)=4,y^'(0)= 1/4 is y=-e^t+5e^{t/4}

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4y^{''}-5y^'+y=0,y(0)=4,y^'(0)= 1/4

4 y^{^{''}} - 5 y^{^{'}} + y = 0, y (0) = 4, y^{^{'}} (0) = \frac{1}{4}

y = - e^{t} + 5 e^{\frac{t}{4}}

Solution steps

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

Solve linear ODE:

y = - e^{t} + 5 e^{\frac{t}{4}}

y = - e^{t} + 5 e^{\frac{t}{4}}

2 y^{^{''}} + 10 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 0

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

a x^{^{''}} + 2 x^{^{'}} = 0

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

What is the solution for 4y^{''}-5y^'+y=0,y(0)=4,y^'(0)= 1/4 ?
The solution for 4y^{''}-5y^'+y=0,y(0)=4,y^'(0)= 1/4 is y=-e^t+5e^{t/4}