4.(a)*+ + ( =0.Solve:(b)Solve xy" + (1– x)y' – 2(1+ x)y = e-*(1 - 6x) by the method based onfactorization of operators.

We have, d2ydx2+dydx+dydx3=0 ..i Put dydx=v ⇒d2ydx2=dvdx Now substitute in equation i we get…

Answered: 4. (a) dzy Solve: dx2 + ( = 0. (b)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

4.
(a)
*+ + ( =0.
Solve:
(b)
Solve xy" + (1– x)y' – 2(1+ x)y = e-*(1 - 6x) by the method based on
factorization of operators.

Expert Solution

Step 1

We have,

$\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + {(\frac{d y}{d x})}^{3} = 0 . . (i)$

Put $\frac{d y}{d x} = v$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d v}{d x}$

Now substitute in equation $(i)$ we get

$\frac{d v}{d x} + v + v^{3} = 0 \Rightarrow \frac{d v}{d x} = - v - v^{3} \Rightarrow \frac{1}{- v - v^{3}} \frac{d v}{d x} = 1 \Rightarrow \frac{1}{- v - v^{3}} d v = d x$

Integrating both sides

$- \ln (v (x)) + \frac{1}{2} \ln (v {(x)}^{2} + 1) = x + c \Rightarrow v (x) = \sqrt{- \frac{1}{1 - e^{2 (x + c)}}}, - \sqrt{- \frac{1}{1 - e^{2 (x + c)}}} N o w, \sin c e v = \frac{d y}{d x} y = \int \sqrt{- \frac{1}{1 - e^{2 (x + c)}}} d x, a n d y = - \int \sqrt{- \frac{1}{1 - e^{2 (x + c)}}} d x \Rightarrow y = \tan^{- 1} (\sqrt{e^{2 x + 2 c} - 1}) + C_{1}, - t {an}^{- 1} (\sqrt{e^{2 x + 2 c} - 1}) + C_{2}$