What is the solution for y^{''}-7y^'+2y=0 ?

The solution for y^{''}-7y^'+2y=0 is y=c_{1}e^{((7+sqrt(41))t)/2}+c_{2}e^{((7-sqrt(41))t)/2}

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

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

y = c_{1} e^{\frac{( 7 + 41 ) t}{2}} + c_{2} e^{\frac{( 7 - 41 ) t}{2}}

Solution steps

y^{''} (t) - 7 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 7 + 41 ) t}{2}} + c_{2} e^{\frac{( 7 - 41 ) t}{2}}

y = c_{1} e^{\frac{( 7 + 41 ) t}{2}} + c_{2} e^{\frac{( 7 - 41 ) t}{2}}

\frac{d ^{2} y}{d t ^{2}} + 7 \frac{d y}{d t} + k y = 0

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

x, \frac{d ^{2} x}{d y ^{2}} + 0.1 x = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{'''}} + 9 y^{^{''}} + 19 y^{^{'}} + 11 y = 0

What is the solution for y^{''}-7y^'+2y=0 ?
The solution for y^{''}-7y^'+2y=0 is y=c_{1}e^{((7+sqrt(41))t)/2}+c_{2}e^{((7-sqrt(41))t)/2}