Answer to Question #327696 in Mechanics | Relativity for John

Question

3d^2y/dx^2+dy/dx-14y=0 ; y=(0) ,y'=(0)=-1

Answer 1

$3y''+y'-14y=0,$

$y=c_1e^{-\frac{7x}3}+c_2e^{2x},$

$y(0)=0: c_1=-c_2,$

$y'(0)=-1: c_2=\frac 76 c_1-\frac 12,$

$c_1=\frac3{13},~c_2=-\frac 3{13},$

$y=\frac 3{13}e^{-\frac{7x}3}-\frac 3{13}e^{2x}.$