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

The solution for y^{''}+y^'-4y=0 is y=c_{1}e^{((-1+sqrt(17))t)/2}+c_{2}e^{((-1-sqrt(17))t)/2}

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

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

y = c_{1} e^{\frac{( - 1 + 17 ) t}{2}} + c_{2} e^{\frac{( - 1 - 17 ) t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 1 + 17 ) t}{2}} + c_{2} e^{\frac{( - 1 - 17 ) t}{2}}

y = c_{1} e^{\frac{( - 1 + 17 ) t}{2}} + c_{2} e^{\frac{( - 1 - 17 ) t}{2}}

y^{^{'''''}} + 4 y^{^{''''}} + 4 y^{^{'''}} = 0

2 y^{^{''}} + 8 y^{^{'}} + 52 y = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (1) = e^{3 x}

4 \frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} + 16 y = 0

y^{^{''}} + 2 y^{^{'}} - 4 y = 0

What is the solution for y^{''}+y^'-4y=0 ?
The solution for y^{''}+y^'-4y=0 is y=c_{1}e^{((-1+sqrt(17))t)/2}+c_{2}e^{((-1-sqrt(17))t)/2}