## Calculus: Early Transcendentals 8th Edition

y = $C_{1}e^{\frac{-2+\sqrt {13}}{3}x}$ + $C_{2}e^{\frac{-2-\sqrt {13}}{3}x}$
Question: 3y"+4y'-3y=0 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(3r^{2}+4r-3)e^{rx}$=0 The corresponding characteristic equation is: $3r^{2}+4r-3=0$ Solving the characteristic equation with the abc-formula and complex numbers gives: $\frac{-b\frac{+}{-}\sqrt {b^{2}-4*a*c}}{2*a}$ $\frac{-4\frac{+}{-}\sqrt {(4)^{2}-4*3*-3}}{2*3}$ $r=\frac{-4\frac{+}{-}\sqrt {52}}{6}$ This can also be written as: $r=\frac{-2\frac{+}{-}\sqrt {13}}{3}$ Adding the constants gives the following general solution: y = $C_{1}e^{\frac{-2+\sqrt {13}}{3}x}$ + $C_{2}e^{\frac{-2-\sqrt {13}}{3}x}$