Answer to Question #277144 in Differential Equations for Jess

Question

Find the general solution of non-homogenous linear differential equation with constant coefficients.

(D² - 3D + 2)y = 5e^2x

Answer 1

The auxiliary equation is $m^2-3m+2= 0$

$m^2-m-2m+2=0$

$m(m-1)-2(m-1) = 0$

$(m-1)(m-2) = 0$

$m= 1$ or $2$

The solution to the homogeneous part is $y = Ae^x + Be^{2x}$

Suppose the general solution is $y = A(x)e^x + B(x)e^{2x}$

Then, $y'= A'(x)e^x+A(x)e^x+B'(x)e^{2x} + 2B(x)e^{2x}$

Set $A'(x)e^x + B'(x)e^{2x}= 0 ... (*)$

Then $y'$ becomes $A(x)e^x+2B(x)e^{2x}$

And $y''= A'(x)e^x+A(x)e^x+2B'(x)e^{2x}+4B(x)e^{2x}$

Substituting all into the given DE we have

$A'(x)e^x+A(x)e^x+2B'(x)e^{2x}+4B(x)e^{2x}-3(A(x)e^x+2B(x)e^{2x})+2(A(x)e^x + B(x)e^{2x}) = 5e^{2x}$

Substituting (*) into the above DE and simplifying further we have

$-A'(x)e^x = 5e^{2x}$

$A'(x)= - 5e^x$

$A(x) =- 5e^x$

And hence $B(x) = 5x$

And the general solution of the DE is

$y = -5e^{2x}+5xe^{2x} = 5(x-1)e^{2x}$