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

The solution for y^{''}-y^'-3y=0 is y=c_{1}e^{((1+sqrt(13))t)/2}+c_{2}e^{((1-sqrt(13))t)/2}

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

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

y = c_{1} e^{\frac{( 1 + 13 ) t}{2}} + c_{2} e^{\frac{( 1 - 13 ) t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( 1 + 13 ) t}{2}} + c_{2} e^{\frac{( 1 - 13 ) t}{2}}

y = c_{1} e^{\frac{( 1 + 13 ) t}{2}} + c_{2} e^{\frac{( 1 - 13 ) t}{2}}

f (x) = ln ∣ cot (3 \cdot x) ∣

((3 x^{2} + 3 x + 4)^{^{'}} (3 x^{2} + 6 x)^{^{'}})^{^{'}}

\int e^{t a n (2 x)} sec^{2} (2 x) d x

y = x^{3} - 2 x^{2} + 5, (2, 5)

\int \frac{1}{3 x ^{4}} d x

What is the solution for y^{''}-y^'-3y=0 ?
The solution for y^{''}-y^'-3y=0 is y=c_{1}e^{((1+sqrt(13))t)/2}+c_{2}e^{((1-sqrt(13))t)/2}