Solve dd2ydt2 − 8ddydt + 15y=9te3t with y(0)=5, y′(0)=10

Name: Solve dd2ydt2 − 8ddydt + 15y=9te3t with y(0)=5, y′(0)=10
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve dd2ydt2 − 8ddydt + 15y=9te3t with y(0)=5, y′(0)=10

Given:d2ydt2 − 8dydt + 15y=9te3twith y(0)=5, y′(0)=10The auxiliary equation is given by m2 − 8m + 15=0solving this we get m=5, 3Hence the complimentary function is yc=C1e−3t + C2e−5tNow The P.I. of given differential equation isyp=1D2 − 8D + 159tx3t=1(D − 3)(D − 5)9te3t=e3t1(D + 3 − 3)(D + 3 − 3)9t=e3t1D(D − 2)9t=e3t1−2D(1 − D2)9t= −e3t21D(1 − D2)−1(1 − D)−19t= −e3t21D(1 + D2 + ⋯)9t= −e−3t21D(9t + 92)= −e−3t21D(92t2+ 92t)= −94e−3t(t + t2)Hence the general solution is y=yc + yp=C1v−3t + C2e5t − 92e−3t(t + t2)Now y(0)=0 ⇒ C1 + e2=5⋯(1)Also y′=3C1e−2t + 5C2e5t − 94[3e−3t(1 + 2t)=3e3t(t + t2)]Hence y′(0)=10 ⇒ 3C1 + 5C2 − 94=10 ⇒ 12C1 + 20C2=49 ⋯(2) solving (1) and (2) we getC1=518, C2=−118Hence the solution is y=518e3t − 118e5t − 94e−3t(t + t2)

"Solve dd2ydt2 − 8ddydt + 15y=9te3t with y(0)=5," was answered by our community

nagasenaz

Answered question

2020-12-06

Solve

d \frac{d^{2} y}{d t^{2}} - 8 d \frac{d y}{d t} + 15 y = 9 t e^{3 t} w i t h y (0) = 5, y^{'} (0) = 10

Answer & Explanation

toroztatG

Skilled2020-12-07Added 98 answers

Given:

\frac{d^{2} y}{d t^{2}} - 8 \frac{d y}{d t} + 15 y = 9 t e^{3 t}

with

y (0) = 5, y^{'} (0) = 10

The auxiliary equation is given by

m^{2} - 8 m + 15 = 0

solving this we get

m = 5, 3

Hence the complimentary function is

y_{c} = C_{1} e^{- 3 t} + C_{2} e^{- 5 t}

Now The P.I. of given differential equation is

y_{p} = \frac{1}{D^{2} - 8 D + 15} 9 t x^{3} t = \frac{1}{(D - 3) (D - 5)} 9 t e^{3} t

= e^{3} t \frac{1}{(D + 3 - 3) (D + 3 - 3)} 9 t

= e^{3} t \frac{1}{D (D - 2)} 9 t

= e^{3} t \frac{1}{- 2 D (1 - \frac{D}{2})} 9 t

= - \frac{e^{3 t}}{2} \frac{1}{D} {(1 - \frac{D}{2})}^{- 1} (1 - D)^{- 1} 9 t

= - \frac{e^{3 t}}{2} \frac{1}{D} (1 + \frac{D}{2} + \dots) 9 t

= - \frac{e^{- 3 t}}{2} \frac{1}{D} (9 t + \frac{9}{2})

= - \frac{e^{- 3 t}}{2} \frac{1}{D} (\frac{9}{2} t^{2} + \frac{9}{2} t)

= - \frac{9}{4} e^{- 3} t (t + t^{2})

Hence the general solution is

y = y_{c} + y p = C_{1} v^{- 3 t} + C_{2} e^{5 t} - \frac{9}{2} e^{- 3} t (t + t^{2})

Now

y (0) = 0 \Rightarrow C_{1} + e_{2} = 5 \dots

(1)
Also

y^{'} = 3 C_{1} e^{- 2 t} + 5 C_{2} e^{5 t} - \frac{9}{4} [3 e^{- 3 t} (1 + 2 t) = 3 e^{3 t} (t + t^{2})]

Hence

y^{'} (0) = 10 \Rightarrow 3 C_{1} + 5 C_{2} - \frac{9}{4} = 10 \Rightarrow 12 C_{1} + 20 C_{2} = 49 \dots

(2) solving (1) and (2) we get

C_{1} = \frac{51}{8}, C_{2} = \frac{- 11}{8}

Hence the solution is

y = \frac{51}{8} e^{3 t} - \frac{11}{8} e^{5 t} - \frac{9}{4} e^{- 3 t} (t + t^{2})

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