Answer to Question #154229 in Differential Equations for K.kaviya

Question

determine the particular solution of the differential equations y''+9y=e^xcosx, using the method of undetermined coefficient

Answer 1

$y''+9y=e^xcosx$

we will find particular solution in the form:

$y_p=Ae^xcosx+Be^xsinx$

$y_p'=Ae^xcosx-Ae^xsinx+Be^xsinx+Be^xcosx=$

$=(A+B)e^xcosx+(B-A)e^xsinx$

$y_p''=(A+B)e^xcosx-(A+B)e^xsinx+$

$+(B-A)e^xsinx+(B-A)e^xcosx=$

$=2Be^xcosx-2Ae^xsinx$

After substitution into equation:

$2Be^xcosx-2Ae^xsinx+9Ae^xcosx+9Be^xsinx=$

$=(2B+9A)e^xcosx+(9B-2A)e^xsinx=e^xcosx$

We have a system of equations:

$2B+9A=1$

$9B-2A=0$

$A=\frac{9B}{2}$

$2B+9*\frac{9B}{2}=1$

$B=\frac{2}{85}$

$A=\frac{9}{85}$

Answer: particular solution is

$y_p=\frac{9}{85}e^xcosx+\frac{4}{85}e^xsinx$