What is the solution for 2y^{''}+3y^'-y=0 ?

The solution for 2y^{''}+3y^'-y=0 is y=c_{1}e^{((-3+sqrt(17))t)/4}+c_{2}e^{((-3-sqrt(17))t)/4}

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2y^{''}+3y^'-y=0

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

y = c_{1} e^{\frac{( - 3 + 17 ) t}{4}} + c_{2} e^{\frac{( - 3 - 17 ) t}{4}}

Solution steps

2 y^{''} (t) + 3 y^{'} (t) - y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 17 ) t}{4}} + c_{2} e^{\frac{( - 3 - 17 ) t}{4}}

y = c_{1} e^{\frac{( - 3 + 17 ) t}{4}} + c_{2} e^{\frac{( - 3 - 17 ) t}{4}}

x \to 0 lim (\frac{x}{∣ x ∣ + x ^{2}})

\frac{e ^{- 4 s}}{( s - 3 ) ^{2}}

x^{2} - 6

f (x) = x - 9

\int \frac{1}{- x ^{2} - 6 x + 40} d x

What is the solution for 2y^{''}+3y^'-y=0 ?
The solution for 2y^{''}+3y^'-y=0 is y=c_{1}e^{((-3+sqrt(17))t)/4}+c_{2}e^{((-3-sqrt(17))t)/4}