What is the solution for y^{''}-2y^'-3y=9 ?

The solution for y^{''}-2y^'-3y=9 is y=c_{1}e^{3t}+c_{2}e^{-t}-3

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y^{''}-2y^'-3y=9

y^{^{''}} - 2 y^{^{'}} - 3 y = 9

y = c_{1} e^{3 t} + c_{2} e^{- t} - 3

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 9

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - 3

y = c_{1} e^{3 t} + c_{2} e^{- t} - 3

5, x^{2} + 1

\int \frac{x ^{6} - x ^{4} + x ^{2}}{x ^{3}} d x

\int_{\frac{1}{2}}^{1} (x^{- 3} - 10) d x

\frac{d}{d x} (5 sin (x) + 6 cos (x))

x \to 2 lim (100)

What is the solution for y^{''}-2y^'-3y=9 ?
The solution for y^{''}-2y^'-3y=9 is y=c_{1}e^{3t}+c_{2}e^{-t}-3