Calculus: Early Transcendentals 8th Edition

y = $cos(\sqrt 3x) + \sqrt 3sin(\sqrt 3x)$
Question: y"+3y=0 y(0)=1 y'(0)=3 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(r^{2}+3)e^{rx}$=0 The corresponding characteristic equation is: $r^{2}+3=0$ Solving the equation with complex numbers gives: $r^{2}=-3$ $r=\frac{+}{-}\sqrt 3 i$ Adding the constants gives the following general solution: y = $C_{1}cos(\sqrt 3x)$ + $C_{2}sin(\sqrt 3x)$ To solve the initial-value problem, we need to differentiate the general solution. y'=$-\sqrt 3C_{1}sin(\sqrt 3x)$ + $\sqrt 3C_{2}cos(\sqrt 3x)$ Fill in the initial value problems in these general solutions. $1 = C_{1}cos(\sqrt 3*0) + C_{2}sin(\sqrt 3*0)$ --> $C_{1}=1$ $3=-\sqrt 3C_{1}sin(\sqrt 3x) + \sqrt 3C_{2}cos(\sqrt 3x)$ --> $C_{2}=\sqrt 3$ The specific solution is: y = $cos(\sqrt 3x) + \sqrt 3sin(\sqrt 3x)$