solve the Cauchy-Euler Differential Equation by substituting x=e^t x^3 (d^3 y/dx^3 ) + 2x^2 (d^2 y/dx^2 ) + 2y = 10x + ((10)/x)


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Question Number 116105 by ShakaLaka last updated on 30/Sep/20

$s o l v e t h e C a u c h y - E u l e r$ $D i f f e r e n t i a l E q u a t i o n b y$ $s u b s t i t u t i n g x = e^{t}$ $x^{3} \frac{d^{3} y}{{d x}^{3}} + 2 x^{2} \frac{d^{2} y}{{d x}^{2}} + 2 y = 10 x + \frac{10}{x}$
	Answered by TANMAY PANACEA last updated on 01/Oct/20
	$x=e^t →(dx/dt)=e^t (dy/dx)=(dy/dt)×(dt/dx)=(dy/dt)×(1/e^t )→e^t (dy/dx)=(dy/dt)→x(dy/dx)=(dy/dt)← (d/dx)(x(dy/dx))=(d/dt)((dy/dt))×(dt/dx) (x(d^2 y/dx^2 )+(dy/dx))×(dx/dt)=(d^2 y/dt^2 ) x(x(d^2 y/dx^2 )+(dy/dx))=(d^2 y/dt^2 ) x^2 (d^2 y/dx^2 )=(d^2 y/dt^2 )−(dy/dt)← (d/dx)(x^2 (d^2 y/dx^2 ))=(d/dt)((d^2 y/dt^2 )−(dy/dt))×(dt/dx) (x^2 (d^3 y/dx^3 )+2x.(d^2 y/dx^2 ))×(dx/dt)=(d^3 y/dt^3 )−(d^2 y/dt^2 ) x^3 (d^3 y/dx^3 )+2x^2 (d^2 y/dx^2 )=(d^3 y/dt^3 )−(d^2 y/dt^2 ) ★ so (d^3 y/dt^3 )−(d^2 y/dt^2 )+2y=10e^t +10e^(−t) now to be solved C.F m^3 −m^2 +2=0 =m^3 +m^2 −2m^2 −2m+2m+2 =m^2 (m+1) −2m (m+1) +2 (m+1) =(m+1)(m^2 −2m+2) m+1=0 m=−1 m^2 −2m+2=0 m=((2±(√(4−8)))/2)=((2±2i)/2)=1±i so m=−1,1±i C.F y=C_1 e^(−t) +C_2 e^((1+i)t) +C_3 e^((1−i)t) P.I correction pls(in black ) y=((10e^t )/(D^3 −D^2 +2))+((10e^(−t) )/(D^3 −D^2 +2)) [D=(d/dt)] (y/(10))=(e^t /(1−1+2))+(e^(−t) /((D+1)(D^2 −2D+2))) =(e^t /2)+(e^(−t) /((D−1+1){(−1)^2 −2(−1)+2})) =(e^t /2)+(e^(−t) /5)×(1/D) =(e^t /2)+(e^(−t) /5)×t y=C.F+P.I y=C_1 e^(−t) +C_2 e^((1+i)t) +C_3 e^((1−i)t) +[(e^t /2)+(e^(−t) /5)×t]×10 =C_1 x^(−1) +C_2 x^(1+i) +C_3 x^(1−i) +((5x)/1)+((2x^(−1) )/)×lnx$
	$x = e^{t} \to \frac{d x}{d t} = e^{t}$ $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{d y}{d t} \times \frac{1}{e^{t}} \to e^{t} \frac{d y}{d x} = \frac{d y}{d t} \to x \frac{d y}{d x} = \frac{d y}{d t} \leftarrow$ $\frac{d}{d x} (x \frac{d y}{d x}) = \frac{d}{d t} (\frac{d y}{d t}) \times \frac{d t}{d x}$ $(x \frac{d^{2} y}{{d x}^{2}} + \frac{d y}{d x}) \times \frac{d x}{d t} = \frac{d^{2} y}{{d t}^{2}}$ $x (x \frac{d^{2} y}{{d x}^{2}} + \frac{d y}{d x}) = \frac{d^{2} y}{{d t}^{2}}$ $x^{2} \frac{d^{2} y}{{d x}^{2}} = \frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t} \leftarrow$ $\frac{d}{d x} (x^{2} \frac{d^{2} y}{{d x}^{2}}) = \frac{d}{d t} (\frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t}) \times \frac{d t}{d x}$ $(x^{2} \frac{d^{3} y}{{d x}^{3}} + 2 x . \frac{d^{2} y}{{d x}^{2}}) \times \frac{d x}{d t} = \frac{d^{3} y}{{d t}^{3}} - \frac{d^{2} y}{{d t}^{2}}$ $x^{3} \frac{d^{3} y}{{d x}^{3}} + 2 x^{2} \frac{d^{2} y}{{d x}^{2}} = \frac{d^{3} y}{{d t}^{3}} - \frac{d^{2} y}{{d t}^{2}} ★$ $s o$ $\frac{d^{3} y}{{d t}^{3}} - \frac{d^{2} y}{{d t}^{2}} + 2 y = 10 e^{t} + 10 e^{- t}$ $n o w t o b e s o l v e d$ $C . F$ $m^{3} - m^{2} + 2 = 0$ $= m^{3} + m^{2} - 2 m^{2} - 2 m + 2 m + 2$ $= m^{2} (m + 1) - 2 m (m + 1) + 2 (m + 1)$ $= (m + 1) (m^{2} - 2 m + 2)$ $m + 1 = 0 m = - 1$ $m^{2} - 2 m + 2 = 0$ $m = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm 2 i}{2} = 1 \pm i$ $s o m = - 1, 1 \pm i$ $C . F$ $y = C_{1} e^{- t} + C_{2} e^{(1 + i) t} + C_{3} e^{(1 - i) t}$ $P . I$ $c o r r e c t i o n p l s (i n b l a c k)$ $y = \frac{10 e^{t}}{D^{3} - D^{2} + 2} + \frac{10 e^{- t}}{D^{3} - D^{2} + 2} [D = \frac{d}{d t}]$ $\frac{y}{10} = \frac{e^{t}}{1 - 1 + 2} + \frac{e^{- t}}{(D + 1) (D^{2} - 2 D + 2)}$ $= \frac{e^{t}}{2} + \frac{e^{- t}}{(D - 1 + 1) {{(- 1)}^{2} - 2 (- 1) + 2}}$ $= \frac{e^{t}}{2} + \frac{e^{- t}}{5} \times \frac{1}{D}$ $= \frac{e^{t}}{2} + \frac{e^{- t}}{5} \times t$ $y = C . F + P . I$ $y = C_{1} e^{- t} + C_{2} e^{(1 + i) t} + C_{3} e^{(1 - i) t} + [\frac{e^{t}}{2} + \frac{e^{- t}}{5} \times t] \times 10$ $= C_{1} x^{- 1} + C_{2} x^{1 + i} + C_{3} x^{1 - i} + \frac{5 x}{1} + \frac{2 x^{- 1}}{} \times l n x$
	Commented by ShakaLaka last updated on 01/Oct/20

	$s i r w h a t m e t h o d d i d y o u a p p l y$ $f o r ‘ ‘ p a r t i c u l a r s o l u t i o n {(y_{p})}^{″} ?$
	Commented by ShakaLaka last updated on 01/Oct/20

	$s i r c a n w e a p p l y m o t h o d o f$ $u n d e t e r m i n e d c o e f f i t i e n t ? i f$ $y e s t h e n w h a t w i l l w e t a k e f o r$ $y_{p} f o r t h e R . H . S$ $Prime causes double exponent: use braces to clarify$
	Commented by TANMAY PANACEA last updated on 01/Oct/20

	$i f o r g o t t o p u t 10 . . .$ $l a t e r c o r r e c t e d$
	Commented by ShakaLaka last updated on 04/Oct/20

	$o h o k I g o t i t . T h a n k y o u :)$
	Answered by mindispower last updated on 01/Oct/20
	$y(x)=y(e^t )=z(t) x=e^t (dy/dx)=(dz/dt).(dt/dx)=z′(t).e^(−t) (d^2 y/dx^2 )=(z′′(t)e^(−t) −z′(t)e^(−t) ).e^(−t) (d^3 y/dx^3 )=(z′′′e^(−2t) −3z′′e^(−2t) +2z′e^(−2t) )e^(−t) equation equivalente e^(3t) (z′′′e^(−3t) −3z′′e^(−2t) +2z′e^(−3t) )+2e^(2t) (z′′−z′)e^(2t) +2z(t)=10(e^t +e^(−t) )=20ch(t) ⇔z′′′−z′′+2z=20ch(t) homegenius Z′′′−Z′′+2Z=0 X^3 −X^2 +2=0⇒(X+1)(X^2 −2X+2)=0 X∈{−1,1−i,1+i} Z(t)=se^(−t) +e^t (acos(t)+bsin(t)) particular solution z=ae^t +bte^(−t) z′=ae^t +be^(−t) −bte^(−t) z′′=ae^t −2be^(−t) +bte^(−t) z′′′=ae^t +3be^(−t) −bte^(−t) ⇒(ae^t +3be^(−t) −bte^(−t) )−(ae^t −2be^(−t) +bte^(−t) ) +2ae^t +2bte^(−t) =20ch(t) ⇒2ae^t +5be^(−t) =20ch(t) ⇒2a=10,5b=10 a=5,b=2 z(t)=se^(−t) +e^t (acos(t)+bsin(t))+5e^t +2te^(−t) y(x)=z(ln(x)) =(s/x)+x(acos(lnx)+bsin(ln(x))+5x+((2ln(x))/x)$
	$y (x) = y (e^{t}) = z (t)$ $x = e^{t}$ $\frac{d y}{d x} = \frac{d z}{d t} . \frac{d t}{d x} = z^{'} (t) . e^{- t}$ $\frac{d^{2} y}{{d x}^{2}} = (z^{″} (t) e^{- t} - z^{'} (t) e^{- t}) . e^{- t}$ $\frac{d^{3} y}{{d x}^{3}} = (z^{‴} e^{- 2 t} - 3 z^{″} e^{- 2 t} + 2 z^{'} e^{- 2 t}) e^{- t}$ $e q u a t i o n$ $e q u i v a l e n t e$ $e^{3 t} (z^{‴} e^{- 3 t} - 3 z^{″} e^{- 2 t} + 2 z^{'} e^{- 3 t}) + 2 e^{2 t} (z^{″} - z^{'}) e^{2 t}$ $+ 2 z (t) = 10 (e^{t} + e^{- t}) = 20 c h (t)$ $\Leftrightarrow z^{‴} - z^{″} + 2 z = 20 c h (t)$ $h o m e g e n i u s$ $Z^{‴} - Z^{″} + 2 Z = 0$ $X^{3} - X^{2} + 2 = 0 \Rightarrow (X + 1) (X^{2} - 2 X + 2) = 0$ $X \in {- 1, 1 - i, 1 + i}$ $Z (t) = {s e}^{- t} + e^{t} (a c o s (t) + b s i n (t))$ $p a r t i c u l a r s o l u t i o n$ $z = {a e}^{t} + {b t e}^{- t}$ $z^{'} = {a e}^{t} + {b e}^{- t} - {b t e}^{- t}$ $z^{″} = {a e}^{t} - 2 {b e}^{- t} + {b t e}^{- t}$ $z^{‴} = {a e}^{t} + 3 {b e}^{- t} - {b t e}^{- t}$ $\Rightarrow ({a e}^{t} + 3 {b e}^{- t} - {b t e}^{- t}) - ({a e}^{t} - 2 {b e}^{- t} + {b t e}^{- t})$ $+ 2 {a e}^{t} + 2 {b t e}^{- t} = 20 c h (t)$ $\Rightarrow 2 {a e}^{t} + 5 {b e}^{- t} = 20 c h (t)$ $\Rightarrow 2 a = 10, 5 b = 10$ $a = 5, b = 2$ $z (t) = {s e}^{- t} + e^{t} (a c o s (t) + b s i n (t)) + 5 e^{t} + 2 {t e}^{- t}$ $y (x) = z (l n (x))$ $= \frac{s}{x} + x (a c o s (l n x) + b s i n (l n (x)) + 5 x + \frac{2 l n (x)}{x}$
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