# solve the differential equation (D3 + 1)y = cos2 (x/2) + e-x

Vandna srivastava rathor
13 Points
2 years ago
general solution =Complementary Function+Particular integral
for Complementary Function m3+1=0
(m+1)(m2+m+1)=0
m=-1, $\frac{-1\pm \sqrt{3}}{2}$
so cf=   $c_{1}e^{-x}+e^{-x/2} (c_{2}cos\frac{\sqrt{3}}{2}x+c_{3}sin\frac{\sqrt{3}}{2}x)$
for pi  =Particular integral
$(D^{3} + 1)y = cos^2 (x/2) + e^{-x}$
$\frac{1}{D^{3} + 1 }\frac{cos(x) }{2}+\frac{1}{D^{3} + 1 }\frac{1 }{2}e^0x+\frac{1}{D^{3} + 1 }e^{-x}$
$-\frac{1 }{2}\frac{D+1}{(D -1)(D+1) }cos(x)+\frac{1 }{2}+x\frac{1}{3(-1)^{2} }e^{-x}$
$-\frac{1 }{2}\frac{D+1}{(D^2 -1) }cos(x)+\frac{1 }{2}+x\frac{1}{3}e^{-x}$
$-\frac{1 }{2}\frac{D+1}{(-1 -1) }cos(x)+\frac{1 }{2}+x\frac{1}{3}e^{-x}$
$\frac{1 }{4}(D+1)cos(x)+\frac{1 }{2}+x\frac{1}{3}e^{-x}$
$\frac{1}{4}(-sinx+cosx)+\frac{1 }{2}+x\frac{1}{3}e^{-x}$
T.S=       cf+pi =$c_{1}e^{-x}+e^{-x/2} (c_{2}cos\frac{\sqrt{3}}{2}x+c_{3}sin\frac{\sqrt{3}}{2}x)+\frac{1}{4}(cosx-sinx)+\frac{1 }{2}+x\frac{1}{3}e^{-x}$