Answer to Question #291609 in Differential Equations for ishita

Question

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

Answer 1

Characteristic (auxiliary) equation

$$3r^2+r-14=0$$

$D=(1)^2-4(3)(-14)=169$

$$r=\dfrac{-1\pm13}{2(3)}$$

$$r_1=-\dfrac{7}{3}, r_2=2$$

The general solution of the differential equation is

$$y=c_1e^{-7x/3}+c_2e^{2x}$$

Given $y(0)=1$

$$1=c_1+c_2$$

$$y'=-\dfrac{7}{3}c_1e^{-7x/3}+2c_2e^{2x}$$

Given $y'(0)=-1$

$$-\dfrac{7}{3}c_1+2c_2=-1$$

$$c_1=\dfrac{9}{13}, c_2=\dfrac{4}{13}$$

The solution of given IVP is

$$y=\dfrac{9}{13}e^{-7x/3}+\dfrac{4}{13}e^{2x}$$