Answer to Question #302575 in Differential Equations for haru

Question

Solve the following initial value problems using method of undetermined coefficients: (i) y''−9y = ex+x−1, y(0) = −1,y'(0) = 1,

Answer 1

Correponding homogeneous differential equation

$y''-9y=0$

Characteristic (auxiliary) equation

$r^2-9=0$

$r_1=3, r_2=-3$

The general solution of the homogeneous differential equation is

$y_h=c_1e^{3x}+c_2e^{-3x}$

Find the particular solution of the non homogeneous differential equation

$y_p=Ae^x+Bx+C$

$y_p'=Ae^x+B$

$y_p''=Ae^x$

Substitute

$Ae^x-9Ae^x-9Bx-9C=e^x+x-1$

$A=-1/8$

$B=-1/9$

$C=1/9$

The general solution of the non homogeneous differential equation is

$y=c_1e^{3x}+c_2e^{-3x}-\dfrac{1}{8}e^x-\dfrac{1}{9}x+\dfrac{1}{9}$

$y'(0) = −1,y(0) = 1$

$3c_1-3c_2-\dfrac{1}{8}-\dfrac{1}{9}=-1$

$c_1+c_2-\dfrac{1}{8}+\dfrac{1}{9}=1$

$c_2=\dfrac{73}{72}-c_1$

$6c_1-\dfrac{219}{72}=-\dfrac{55}{72}$

$c_1=\dfrac{82}{216}$

$c_2=\dfrac{137}{216}$

The solution of the given IVP is

$y=\dfrac{82}{216}e^{3x}+\dfrac{137}{216}e^{-3x}-\dfrac{1}{8}e^x-\dfrac{1}{9}x+\dfrac{1}{9}$