Answer to Question #193734 in Differential Equations for YASH YADAV

Question

Solve (D^2 -4D+3)y = e^x cos 2x + cos3x 

Answer 1

We have given the differential equation,

$(D^2 -4D+3)y = e^x cos 2x + cos3x$

Auxiliary equation can be given as,

$m^2-4m+3 =0\\ m^2-3m-m+3 = 0\\ m(m-3)-1(m-3) = 0\\ m=1,2$  

Hence, $CF = C_1e^x+C_2e^{2x}$

Particular integral can be calculated as,

$PI = \dfrac{cos3x}{-9-4D+3}+ \dfrac{e^xcos2x}{(D+1)^2-4(D+1)+3}$

We know,

$\dfrac{1}{f(D^2)}sinax = \dfrac{sinax}{f(-a^2)}$ and $\dfrac{1}{f(D^2)}cosax = \dfrac{cosax}{f(-a^2)}$

Now Multiplying by $(D-3)$ in numerator and denominator of first part of particular integral.

$PI = \dfrac{cos3x(D-3)}{-2(D^2-9)}+ \dfrac{e^xcos2x}{D^2-2D}$

After solving we get,

$PI = \dfrac{-3sin3x-3cos3x}{36} + \dfrac{e^x(-2sin2x-2cos2x)}{16}$

Hence , the complete solution is

$y = C_1e^x+C_2e^{2x}+ \dfrac{-3sin3x-3cos3x}{36} + \dfrac{e^x(-2sin2x-2cos2x)}{16}$