Find the general solution of the given differential equation.&nbsp;y''+2y'=3+4sin2t

First find solution of homogeneous problem.r2+2r=0⇒r1,2=0,−2Homogeneous solution:yc=c1+c2e−2tLet&nbsp;Y=At+Bcos2t+Csin2t&nbsp;- because&nbsp;g(t)=3+4sin2tPlug Y into starting equation to find particular solution:(At+Bcos2t+Csin2t)''+2(At+Bcos2t+Csin2t)'=3+4sin2t-4Bcos2t-4Csin2t+2A-4Bsin2t+4Ccos2t=3+4sin2t2A+cos2t(-4B+4C)+sin2t(-4C-4B)=3+4sin2tWe obtain system:2A=3⇒A=32-4C-4B=4-4B+4C=0&nbsp;- add equations−8B=4⇒B=−12⇒C=−12Y=32t-12cos2t-12sin2tSolution of problem:y=yc+Y=c1+c2e-2t+32t-12cos2t-12sin2tResult:y=c1+c2e-2t+32t-12cos2t-12sin2t

"Find the general solution of the given differential..." was answered by our experts

Johan Patton

Open question

2022-08-19

Find the general solution of the given differential equation. $y'' + 2 y' = 3 + 4 \sin 2 t$

Answer & Explanation

Ben Reese

Beginner2022-08-20Added 12 answers

First find solution of homogeneous problem.
$r^{2} + 2 r = 0 \Rightarrow r_{1, 2} = 0, - 2$
Homogeneous solution:
$y_{c} = c_{1} + c_{2} e^{- 2 t}$
Let $Y = A t + B \cos 2 t + C \sin 2 t$ - because $g (t) = 3 + 4 \sin 2 t$
Plug Y into starting equation to find particular solution:
$(A t + B \cos 2 t + C \sin 2 t)'' + 2 (A t + B \cos 2 t + C \sin 2 t)' = 3 + 4 \sin 2 t$
$- 4 B \cos 2 t - 4 C \sin 2 t + 2 A - 4 B \sin 2 t + 4 C \cos 2 t = 3 + 4 \sin 2 t$
$2 A + \cos 2 t (- 4 B + 4 C) + \sin 2 t (- 4 C - 4 B) = 3 + 4 \sin 2 t$
We obtain system:
$2 A = 3 \Rightarrow A = \frac{3}{2}$
$- 4 C - 4 B = 4$
$- 4 B + 4 C = 0$ - add equations
$- 8 B = 4 \Rightarrow B = - \frac{1}{2} \Rightarrow C = - \frac{1}{2}$
$Y = \frac{3}{2} t - \frac{1}{2} \cos 2 t - \frac{1}{2} \sin 2 t$
Solution of problem:
$y = y_{c} + Y = c_{1} + c_{2} e^{- 2 t} + \frac{3}{2} t - \frac{1}{2} \cos 2 t - \frac{1}{2} \sin 2 t$
Result:
$y = c_{1} + c_{2} e^{- 2 t} + \frac{3}{2} t - \frac{1}{2} \cos 2 t - \frac{1}{2} \sin 2 t$

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