Answer
Answer attached below.
Work Step by Step
$y''-(a+b)y'+aby=0$ _____($\alpha$)
$y=C_{1} e^{ax}+C_{2}e^{bx}$ ______(1)
Differentiate (1) with respect to $x$
$y'=aC_{1} e^{ax}+bC_{2}e^{bx}$ ______(2)
Differentiate (2) with respect to $x$
$y''=a^2C_{1} e^{ax}+b^2C_{2}e^{bx}$ ______(3)
$-(a+b)y'=-(a+b)[aC_{1} e^{ax}+bC_{2}e^{bx}]$
$-(a+b)y'= -a^2C_{1}e^{ax}-abC_{2}e^{bx}-abC_{1}e^{ax}-b^2C_{2}e^{bx}$ ____(4)
$aby=abC_{1} e^{ax}+abC_{2}e^{bx}$ ____(5)
Adding (3),(4) and (5)
$y''-(a+b)y'+aby =a^2C_{1} e^{ax}+b^2C_{2}e^{bx}-a^2C_{1}e^{ax}-abC_{2}e^{bx}-abC_{1}e^{ax}-b^2C_{2}e^{bx}+abC_{1} e^{ax}+abC_{2}e^{bx} =0$
Hence, $y=C_{1} e^{ax}+C_{2}e^{bx}$ is solution of ($\alpha$)
