Answer to Question #218211 in Differential Equations for Jck

Question

Find y as a function of t

 if 16y′′ + 136y′ +289y = 0,

y(0) = 7 , y′(0) = 3.

Answer 1

$$16y'' + 136y' +289y = 0$$

The characteristic equation

$$16r^2+136r+289=0$$

$$(4r+17)^2=0$$

$$r_{1,2}=-\dfrac{17}{4}$$

The general solution of the homogeneous equation is

$$y(t)=c_1e^{(-17/4)t}+c_2te^{(-17/4)t}$$

$$y(0)=7:c_1e^{(-17/4)(0)}+c_2(0)e^{(-17/4)(0)}=7$$

$$c_1=7$$

$$y=7e^{(-17/4)t}+c_2te^{(-17/4)t}$$

$$y'=-\dfrac{119}{4}e^{(-17/4)t}+c_2e^{(-17/4)t}-\dfrac{17}{4}c_2te^{(-17/4)t}$$

$$y'(0)=3:-\dfrac{119}{4}+c_2-0=3$$

$$c_2=\dfrac{131}{4}$$

The solution of the given Initial Value Problem is

$$y(t)=7e^{(-17/4)t}+\dfrac{131}{4}te^{(-17/4)t}$$