Find a suitable form for particular solution of the differential equation y′′+4y=e−2t(t2+1)

Question

Nathaniel Kramer · Accepted Answer

Solve d2y(t)dt2+4y(t)=−2t(t2+1)+e: The general solution will be the sum of the complementary solution and particular solution. Find the complementary solution by solving d2y(t)dt2+4y(t)=0: Assume a solution will be proportional to eλt for some constant λ. Substitute y(t)=eλt into the differential equation: d2dt2(eλt)+4eλt=0 Substitute d2dt2(eλt)=λ2eλt: λ2{e}{λ{t}}+{4}{e}{λ{t}}={0} Factor out eλt: (λ2+4)eλt=0 Since eλt≠0 for any finite λ, the zeros must come from the polynomial: λ2+4=0 Solve for λ: λ=2i or λ=−2i The roots λ=±2i give y1(t)=c1e2it,y2(t)=c2e−2it as solutions, where c1andc2 are arbitrary constants. The general solution is the sum of the above solutions: y(t)=y1(t)+y2(t)=c1e2it+c2e−2it Apply Euler&#039;s identity eα+iβ=eαcos⁡(β)+ieαsin⁡(β): y(t)=c1(cos⁡(2t)+isin⁡(2t))+c2(cos⁡(2t)−isin⁡(2t)) Regroup terms: y(t)=(c1+c2)cos⁡(2t)+i(c1−c

Find a suitable form for particular solution of the differential equation . y′′+4y=e−2t(t2+1)

Answered question

Answer & Explanation

New Questions in Calculus 1