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

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

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

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

y = (1 - \frac{5}{4}) e^{4 t} + \frac{5}{4}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{5}{4}) e^{4 t} + \frac{5}{4}

y = (1 - \frac{5}{4}) e^{4 t} + \frac{5}{4}

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

x^{^{''}} + \frac{15}{\frac{1}{32}} x^{^{'}} + \frac{13}{\frac{1}{32}} x = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 4, y^{^{'}} (- 1) = 2

(D^{3} - 5 D^{2} + 8 D - 4) y = 0

y^{^{''}} + b \cdot y = 0, y (0) - 9 y^{^{'}} (0) = 0, y^{^{'}} (9) = 0

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