What is the solution for 4y^{''}-21y^'+20y=0,y^'(0)=-2.75 ?

The solution for 4y^{''}-21y^'+20y=0,y^'(0)=-2.75 is y=(e^{4t}(-2.75-1.25c_{2}))/4+c_{2}e^{(5t)/4}

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4y^{''}-21y^'+20y=0,y^'(0)=-2.75

4 y^{^{''}} - 21 y^{^{'}} + 20 y = 0, y^{^{'}} (0) = - 2.75

y = \frac{e ^{4 t} ( - 2.75 - 1.25 c _{2} )}{4} + c_{2} e^{\frac{5 t}{4}}

Solution steps

4 y^{''} (t) - 21 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = \frac{e ^{4 t} ( - 2.75 - 1.25 c _{2} )}{4} + c_{2} e^{\frac{5 t}{4}}

y = \frac{e ^{4 t} ( - 2.75 - 1.25 c _{2} )}{4} + c_{2} e^{\frac{5 t}{4}}

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = e^{2 x}

x^{^{''}} + 15 x^{^{'}} + 24 x = 0

3 y^{^{''}} - 12 y^{^{'}} + 87 y = 0

x^{^{''}} - 2 x^{^{'}} - x = 0

y^{^{'''}} + 3 y^{^{''}} - 28 y^{^{'}} = 0

What is the solution for 4y^{''}-21y^'+20y=0,y^'(0)=-2.75 ?
The solution for 4y^{''}-21y^'+20y=0,y^'(0)=-2.75 is y=(e^{4t}(-2.75-1.25c_{2}))/4+c_{2}e^{(5t)/4}