1.

Find:(D2 + 3D + 2)y = e2x-sin x

Answer»

(D2 + 3D + 2)y = e2x-sin x

m2 + 3m + 2 = 0 is its auxiliary equation

⇒ (m+ 1)(m + 2) = 0

⇒ m = -1, -2

∴ C.F. = C1e-x + C2e-2x

P.I. = \(\frac1{D^2+3D+2}e^{2x}-\frac1{D^2+3D+2}sin x\)

 = \(\frac{e^{2x}}{12}-\frac{sin x}{-1+3D+2}\)

 = \(\frac{e^{2x}}{12}-\frac{sin x}{3D+1}\) 

 = \(\frac{e^{2x}}{12}-\frac{(3D-1)}{9D^2-1}sin x\) 

 = \(\frac{e^{2x}}{12}-\frac{(3D-1)sin x}{-9-1}\) 

 = \(\frac{e^{2x}}{12}-\frac{3cos x-sin x}{-10}\) 

 = \(\frac{e^{2x}}{12}+\frac3{10}cos x-\frac3{10} sin x\)



Discussion

No Comment Found

Related InterviewSolutions