y′′′ +4y′ = 3x−1 what is solution


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 81554 by jagoll last updated on 14/Feb/20

$y^{‴} + 4 y^{'} = 3 x - 1$ $w h a t i s s o l u t i o n$
	Answered by TANMAY PANACEA last updated on 14/Feb/20
	$let y=e^(mx) be a solution so we get y^′ =(dy/dx)=me^(mx) →y^(′′) =m^2 e^(mx) y^(′′′) =m^3 e^(mx) complemenary function m^3 e^(mx) +4me^(mx) =0 e^(mx) ×m(m^2 +4)=0 e^(mx) ≠0 so eithethe m=0 or m=2i or −2i C.F=Ae^(0.x) +Be^(2ix) +Ce^(−2ix) (A,B and C constant) now calculation of particular intregal y=((3x−1)/(D^3 +4D)) where D=(d/dx) y=(1/D).(1/(D^2 +4)).(3x−1) y=(4+D^2 )^(−1) ×(((3x^2 )/2)−x) y={4(1+(D^2 /4))}^(−1) ×(((3x^2 )/2)−x) =(1/4)(1−(D^2 /4)+(D^4 /(16))+...)(((3x^2 )/2)−x) note (1+p)^(−1) =1−p+p^2 −p^3 +...∞ calculation (1/4)(((3x^2 )/2)−x)−(D^2 /(16))(((3x^2 )/2)−x)+(D^4 /(64))(((3x^2 )/2)−x)..othester is zero ((3x^2 )/8)−(x/4)−(3/(32))×2 =((3x^2 )/8)−(x/4)−(3/(16)) so compete solution y=Ae^(0.x) +Be^(2ix) +Ce^(−2ix) +((3x^2 )/8)−(x/4)−(3/(16)) note..(1)we can put e^(ix) =cosx+isinx e^(−ix) =cosx−isinx and (A−(3/4))=K new constant pls check$
	$l e t y = e^{m x} b e a s o l u t i o n s o w e g e t$ $y^{^{'}} = \frac{d y}{d x} = {m e}^{m x} \to y^{^{″}} = m^{2} e^{m x}$ $y^{^{‴}} = m^{3} e^{m x}$ $c o m p l e m e n a r y f u n c t i o n$ $m^{3} e^{m x} + 4 {m e}^{m x} = 0$ $e^{m x} \times m (m^{2} + 4) = 0$ $e^{m x} \neq 0 s o$ $e i t h e t h e m = 0 o r m = 2 i o r - 2 i$ $C . F = {A e}^{0 . x} + B e^{2 i x} + C e^{- 2 i x} (A, B a n d C c o n s t a n t)$ $n o w$ $c a l c u l a t i o n o f p a r t i c u l a r i n t r e g a l$ $y = \frac{3 x - 1}{D^{3} + 4 D} w h e r e D = \frac{d}{d x}$ $y = \frac{1}{D} . \frac{1}{D^{2} + 4} . (3 x - 1)$ $y = {(4 + D^{2})}^{- 1} \times (\frac{3 x^{2}}{2} - x)$ $y = {4 (1 + \frac{D^{2}}{4})}^{- 1} \times (\frac{3 x^{2}}{2} - x)$ $= \frac{1}{4} (1 - \frac{D^{2}}{4} + \frac{D^{4}}{16} + . . .) (\frac{3 x^{2}}{2} - x)$ $n o t e$ ${(1 + p)}^{- 1} = 1 - p + p^{2} - p^{3} + . . . \infty$ $c a l c u l a t i o n$ $\frac{1}{4} (\frac{3 x^{2}}{2} - x) - \frac{D^{2}}{16} (\frac{3 x^{2}}{2} - x) + \frac{D^{4}}{64} (\frac{3 x^{2}}{2} - x) . . o t h e s t e r i s z e r o$ $\frac{3 x^{2}}{8} - \frac{x}{4} - \frac{3}{32} \times 2$ $= \frac{3 x^{2}}{8} - \frac{x}{4} - \frac{3}{16}$ $s o c o m p e t e s o l u t i o n$ $y = {A e}^{0 . x} + {B e}^{2 i x} + {C e}^{- 2 i x} + \frac{3 x^{2}}{8} - \frac{x}{4} - \frac{3}{16}$ $n o t e . . (1) w e c a n p u t e^{i x} = c o s x + i s i n x$ $e^{- i x} = c o s x - i s i n x$ $a n d (A - \frac{3}{4}) = K n e w c o n s t a n t$ $p l s c h e c k$
	Commented by jagoll last updated on 14/Feb/20

	$c a n w e g e t g r a d e s A, B a n d C m i s t e r ?$
	Commented by mr W last updated on 14/Feb/20

	$y o u g a v e o n l y a d . e . w i t h o u t i n i t i a l$ $c o n d i t i o n s, s o y o u c a n n o t d e t e r m i n e$ $t h e s e c o n s t a n t s . t o d e t e r m i n e A, B, C$ $y o u n e e d t h r e e i n i t i a l c o n d i t i o n s s u c h a s$ $y (0) = . . .$ $y^{'} (1) = . . .$ $y^{″} (2) = . . .$
	Commented by jagoll last updated on 14/Feb/20

	$o o y e s s i r t h a n k y o u$
	Answered by Joel578 last updated on 14/Feb/20
	$• homogeneous solution with characteristic eq. λ^3 + 4λ = 0 ⇒ λ_1 = 0, λ_(2,3) = ± 2i ⇒ y_h (x) = C_1 + C_2 cos 2x + C_3 sin 2x • particular solution y_p = Ax^2 + Bx + C y_p ′ = 2Ax + B y_p ′′ = 2A y_p ′′′ = 0 Substitute to original question 0 + 4(2Ax + B) = 3x − 1 ⇒ { ((8A = 3)),((4B = −1)) :} ⇒ { ((A = (3/8))),((B = ((−1)/4))) :} ⇒ y_p ′ = (3/4)x − (1/4) ⇒ y_p = (3/8)x^2 − (1/4)x + K ∴ y(x) = y_h (x) + y_p (x) = C + C_2 cos 2x + C_3 sin 2x + (3/8)x^2 − (1/4)x$
	$∙ homogeneous solution with characteristic eq .$ $λ^{3} + 4 λ = 0 \Rightarrow λ_{1} = 0, λ_{2, 3} = \pm 2 i$ $\Rightarrow y_{h} (x) = C_{1} + C_{2} \cos 2 x + C_{3} \sin 2 x$ $∙ particular solution$ $y_{p} = {A x}^{2} + B x + C$ $y_{p}^{'} = 2 A x + B$ $y_{p}^{″} = 2 A$ $y_{p}^{‴} = 0$ $Substitute to original question$ $0 + 4 (2 A x + B) = 3 x - 1$ $\Rightarrow {\begin{cases} 8 A = 3 \\ 4 B = - 1 \end{cases} \Rightarrow {\begin{cases} A = \frac{3}{8} \\ B = \frac{- 1}{4} \end{cases}$ $\Rightarrow y_{p}^{'} = \frac{3}{4} x - \frac{1}{4} \Rightarrow y_{p} = \frac{3}{8} x^{2} - \frac{1}{4} x + K$ $∴ y (x) = y_{h} (x) + y_{p} (x)$ $= C + C_{2} \cos 2 x + C_{3} \sin 2 x + \frac{3}{8} x^{2} - \frac{1}{4} x$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum