Question

Show that , for the DE
d²y/dx² + a(x) dy/dx+b(x)y=0

exp(mx) is a particular integral if m²+am+b=0 . Hence find the value of m so that exp(mx) is a particular integral of the equation
(x-2) d²y/dx² - (4x-7)dy/dx+(4x-6)y=0

Answer 1

Let's make a substitution $y=exp(mx)$

$y'=my$

$y''=m^2y$

Hence

$y''+a(x)y'+b(x)y=m^2y+a(x)my+b(x)y$

Therefore, if

$m^2+am+b=0$

$y''+a(x)y'+b(x)y=0$

Which means y=exp(mx) is particular integral of the equation.  

In the equation $(x-2)y''-(4x-7)y'+(4x-6)y=0$

$a=-\frac{4x-7}{x-2};\quad b=\frac{4x-6}{x-2}$

Let's solve the eqation

$m^2+am+b=0$

$m^2-\frac{4x-7}{x-2}m+\frac{4x-6}{x-2}=0$

$m=\frac{4x-7-\sqrt{(4x-7)^2-4(x-2)(4x-6)}}{2(x-2)}=\frac{4x-7-1}{2(x-2)}=2$

therefore $y=e^{2x}$ is a particular integral of the equation