Q16E In Problems 11鈥�18, find a gene... [FREE SOLUTION]

91影视

Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 4: Q16E (page 191)

In Problems 11鈥�18, find a general solution to the differential equation.
$y'' + 5 y' + 6 y = 18 t^{2}$

Short Answer

Expert verified

The general solution is $y (t) = c_{1} e^{- 3 t} + c_{2} e^{- 2 t} + 3 t^{2} - 5 t + \frac{19}{6}$ .

Step by step solution

Find a particular solution

The auxiliary equation is $r^{2} + 5 r + 6 = 0$ .

Two independent solutions are $r = - 2, - 3$ .

Then $y_{1} = e^{- 3 t}, v_{2} = e^{- 2 t}$

$y_{h} (t) = c_{1} e^{- 3 t} + c_{2} e^{- 2 t}$

The particular solution is $y_{p} = v_{1} (t) e^{- 3 t} + v_{2} (t) e^{- 2 t}$ .

Find v1' and v1

Now,

$\begin{array}{rcl} v_{1}' & = & \frac{- f (t) y_{2} (t)}{a [y_{1} (t) y'_{2} (t) - y'_{1} (t) y_{2} (t)]} \\ = & \frac{- 18 t^{2} . e^{- 2 t}}{[e^{- 3 t} (- 2 e^{- 2 t}) + 3 e^{- 3 t} e^{- 2 t}]} \\ = & \frac{- 18 t^{2} . e^{- 2 t}}{[- 2 e^{- 5 t} + 3 e^{- 5 t}]} \\ = & \frac{- 18 t^{2} . e^{- 2 t}}{[e^{- 5 t}]} \\ = & - 18 t^{2} e^{3 t} \end{array}$

Now integrating this by integration by parts formula

$\begin{array}{rcl} v_{1} (t) & = & 鈭� - 18 t^{2} e^{3 t} d t \\ = & \frac{- 2 e^{3 t}}{3} (2 - 6 t + 9 t^{2}) + C \end{array}$

Determine v2' and v2

$\begin{array}{rcl} v_{2}' & = & \frac{f (t) y_{1} (t)}{a [y_{1} (t) y'_{2} (t) - y'_{1} (t) y_{2} (t)]} \\ = & \frac{18 t^{2} . e^{- 3 t}}{[e^{- 3 t} (- 2 e^{- 2 t}) + 3 e^{- 3 t} e^{- 2 t}]} \\ = & 18 t^{2} e^{2 t} \end{array}$

Integrate this by integration by parts.

$\begin{array}{rcl} y_{2} (t) & = & 鈭� 18 t^{2} e^{2 t} d t \\ = & \frac{9}{2} e^{2 t} (1 - 2 t + 2 t^{2}) + C \end{array}$

Thus, a particular solution is:

role="math" localid="1664254217788" $\begin{array}{rcl} y_{p} & = & (\frac{- 2 e^{3 t}}{3} (2 - 6 t + 9 t^{2}) + C) e^{- 3 t} + (\frac{9}{2} e^{2 t} (1 - 2 t + 2 t^{2}) + C) e^{- 2 t} \\ y_{p} & = & 3 t^{2} - 5 t + \frac{19}{6} \end{array}$

Therefore, the general solution is:

$y (t) = y_{h} (t) + y_{p} (t) y (t) = c_{1} e^{- 3 t} + c_{2} e^{- 2 t} + 3 t^{2} - 5 t + \frac{19}{6}$

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91影视

In Problems 11鈥�18, find a general solution to the differential equation.
$y'' + 5 y' + 6 y = 18 t^{2}$

Short Answer

Step by step solution

Find a particular solution

Find v1' and v1

Determine v2' and v2

One App. One Place for Learning.

Most popular questions from this chapter

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91影视

In Problems 11鈥�18, find a general solution to the differential equation.y''+5y'+6y=18t2

Short Answer

Step by step solution

Find a particular solution

Find v1' and v1

Determine v2' and v2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Geometry

Calculus

Discrete Mathematics

Mechanics Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

In Problems 11鈥�18, find a general solution to the differential equation.
$y'' + 5 y' + 6 y = 18 t^{2}$