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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: Q48 E (page 201)

Complete the proof of Theorem 8by solving equation (16).

Short Answer

Expert verified

The proof of Theorem 8 is $y_{2} = {vy}_{1} = y_{1} 鈭� \frac{- e^{鈭� p (t) dt}}{y_{1}^{2}} dt$ by solving equation (16).

Step by step solution

01

Using the Equation

From (16) we have $y_{1} w' + ({2 y}_{1}^{}' + {py}_{1}) w = 0$ then

$y_{1} w' = - 2 (y_{1}^{}' + {py}_{1}) w \frac{w'}{w} = - \frac{2 (y_{1}^{}' + {py}_{1})}{y_{1}}$

Integrate on both sides.

$鈭� \frac{w'}{w} dw = - 鈭� \frac{2 (y_{1}^{}' + {py}_{1})}{y_{1}} dt 鈭� \frac{w'}{w} dw = - 鈭� (2 \frac{y_{1}^{}'}{y_{1}} + p) dt - lnw = 2 {lny}_{1} + 鈭� p (t) dtw = e^{- (2 {lny}_{1} + 鈭� p (t) dt} w = \frac{- e^{鈭� p (t) dt}}{y_{1}^{2}}$

02

Substitute the value for "w"

One has $w = v'$ ,Therefore,

$\frac{dv}{dt} = \frac{- e^{鈭� p (t) dt}}{y_{1}^{2}} dv = \frac{- e^{鈭� p (t) dt}}{y_{1}^{2}} dt v = 鈭� \frac{- e^{鈭� p (t) dt}}{y_{1}^{2}} dt$

And

$y_{2} = {vy}_{1} = y_{1} 鈭� \frac{- e^{鈭� p (t) dt}}{y_{1}^{2}} dt$

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