Solve(d2ydt2) + 7(dydt) + 10y=4te−3t with y(0)=0, y′(0)= −1

Name: Solve(d2ydt2) + 7(dydt) + 10y=4te−3t with y(0)=0, y′(0)= −1
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve(d2ydt2) + 7(dydt) + 10y=4te−3t with y(0)=0, y′(0)= −1

Given:d2ydt2 + 7dydt + 10y=4te−3twith y(0)=0, y′(0)= −1The auxiliary equation is given by m2 + 7m + 10=0solving this we get m= −5, −2Hence the complimentary function is yc=C1e2t + C2e−5tNow The P.I. of given differential equation isyp=1D2 + 7D + 104tx−3t=1(d + 5)(D + 2)4te−3t=e3t1(D − 3 + 5)(D − 3 + 2)4t=e3t1(D + 2)(D − 1)4t=e3t1−2(1 + D2)(1 − D)4t= −e−3t2(1 + D2)1(1 − D)−14t= −e−3t2(1 − D6 + ⋯)(1 + D + D2)4t= −e−3t2(1 + (1 − 12)D + ⋯)4t= −e−3t2(4t + 2)= −e−3t(2t + 1)Hence the general solution is y=yc+yp=C1v−2t+C2e5t−e−3t(2t+1)Now y(0)=0 ⇒ C1 + e2 − 1=0 ⇒ C1=C2=1 ⋯(1)Also y′= −2C1e−2t − 5C2e−5t + 3e−3t(2t + 1) − 2e−3tHence y′(0)= −1⇒ −2C1 − 5C2 − 2 + 3= −1 ⇒ 2C1 + 5C2=2 ⋯ (2)solving (1) and (2) we get C1=1, C2=0Hence the solution is y=e−2t − e−3t(2t + 1)

Solved: "Solve(d2ydt2) + 7(dydt) + 10y=4te−3t with y(0)=0, y′(0)="

Braxton Pugh

Answered question

2021-02-15

Solve

(d^{2} \frac{y}{d t^{2}}) + 7 (\frac{d y}{d t}) + 10 y = 4 t e^{- 3} t

with

y (0) = 0, y^{'} (0) = - 1

Answer & Explanation

Benedict

Skilled2021-02-16Added 108 answers

Given:
$\frac{d^{2} y}{d t^{2}} + 7 \frac{d y}{d t} + 10 y = 4 t e^{- 3} t$
with $y (0) = 0, y^{'} (0) = - 1$
The auxiliary equation is given by $m^{2} + 7 m + 10 = 0$
solving this we get $m = - 5, - 2$
Hence the complimentary function is $y_{c} = C_{1} e^{2 t} + C_{2} e^{- 5} t$
Now The P.I. of given differential equation is
$y_{p} = \frac{1}{D^{2} + 7 D + 10} 4 t x^{- 3} t = \frac{1}{(d + 5) (D + 2)} 4 t e^{- 3} t$
$= e^{3} t \frac{1}{(D - 3 + 5) (D - 3 + 2)} 4 t$
$= e^{3} t \frac{1}{(D + 2) (D - 1)} 4 t$
$= e^{3} t \frac{1}{- 2 (1 + \frac{D}{2}) (1 - D)} 4 t$
$= - \frac{e^{- 3} t}{2} {(1 + \frac{D}{2})}^{1} (1 - D)^{- 1} 4 t$
$= - \frac{e^{- 3} t}{2} (1 - \frac{D}{6} + \dots) (1 + D + D^{2}) 4 t$
$= - \frac{e^{- 3} t}{2} (1 + (1 - \frac{1}{2}) D + \dots) 4 t$
$= - \frac{e^{- 3} t}{2} (4 t + 2)$
$= - e^{- 3} t (2 t + 1)$
Hence the general solution is $y = y_{c} + y_{p} = C_{1} v^{- 2} t + C_{2} e^{5} t - e^{- 3} t (2 t + 1)$
Now $y (0) = 0 \Rightarrow C_{1} + e_{2} - 1 = 0 \Rightarrow C_{1} = C_{2} = 1 \dots$ (1)
Also $y^{'} = - 2 C_{1} e^{- 2} t - 5 C_{2} e^{- 5} t + 3 e^{- 3} t (2 t + 1) - 2 e^{- 3} t$
Hence $y^{'} (0) = - 1 \Rightarrow - 2 C_{1} - 5 C_{2} - 2 + 3 = - 1 \Rightarrow 2 C_{1} + 5 C_{2} = 2 \dots$ (2)
solving (1) and (2) we get $C_{1} = 1, C_{2} = 0$
Hence the solution is $y = e^{- 2} t - e^{- 3} t (2 t + 1)$

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