Solve left(d^{2}frac{y}{dt^{2}}right) + 7left(frac{dy}{dt}right) + 10y=4te^{-3}t with y(0)=0, y'(0)= -1

Question

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 1

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 $y = e^{- 2} t - e^{- 3} t (2 t + 1)$

Solve left(d^{2}frac{y}{dt^{2}}right) + 7left(frac{dy}{dt}right) + 10y=4te^{-3}t with y(0)=0, y'(0)= -1

Expert Answer

Not exactly what you’re looking for?

Relevant Questions