Solve the differential equation (36D2−24D+13)y=2sin2x−e−x+2
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Solution
The characteristic equation is 36p2−24p+13=0 ⇒p=−b±√b2−4ac2a =24±√242−4(36)(13)72 (a=36,b=−24,c=13) 24±√−129672=24±26i72=12(2±3i)72 =2±3i6=26+3i6=13+i2
Complementary functions is e13x[Acos12x+Bsinx2]....(1) PI=2sin2x−e−x+2=1−cos2x−e−x+2=3−cos2x−e−x[2.sin2x=1−cos2x] PI1=3.eax36D−24D+13=3.eax0−0+13=313eax=313....(2) PI2=−e−x36D2−24D+13=−e−x36(−1)2−24(−1)+13....(3) =−e−x36+24+13=−e−x73 PI3=−cos2x36D2−24D+13=−cos2x36(−4)−24D+13 =−cos2x−144−24D+13=−cos2x−24D−131 =cos2x24D+131×24D−13124D−131=(24D+131)cos2x576D2−17161 =(24D+131)cos2x576(−4)−17161=(24D−131)cos2x−2304−17161 =(24D−131)cos2x−19465=24(2)(−sin2x)−131cos2x−19465 =−48sin2x−131cos2x−19465 =−48sin2x+131cos2x−19465 ∴ Solution is y=CF+PI1+PI2+PI3 y=ex3[Acosx2+Bsinx2]+313−e−x73+48sin2x+131cos2x19465