Find-all-functions-y-f-x-such-that-y-y-y-

Question Number 55668 by mr W last updated on 01/Mar/19

F i n d a l l f u n c t i o n s y = f (x) s u c h t h a t

y^{'} y^{″} = y^{‴} .

Commented by malwaan last updated on 02/Mar/19

y = e^{x}

Commented by mr W last updated on 02/Mar/19

p l e a s e c h e c k s i r :

w i t h y = e^{x}

y^{'} = e^{x}

y^{″} = e^{x}

y^{‴} = e^{x}

y^{'} y^{″} = e^{x} e^{x} = e^{2 x} \neq y^{^{‴}} = e^{x}

\Rightarrow y = e^{x} i s n o t a s o l u t i o n

Commented by mr W last updated on 02/Mar/19

{i t}^{'} s o b v i o u s t h a t y = c_{1} x + c_{2} i s a s o l u t i o n,

s i n c e y^{'} = c_{1}, y^{″} = 0, y^{‴} = 0 \Rightarrow y^{'} y^{″} = y^{‴} = 0 .

b u t a r e t h e r e a n y o t h e r s o l u t i o n s ?

Answered by mr W last updated on 03/Mar/19

l e t g (x) = y^{'} = f^{'} (x)

\Rightarrow {g g}^{'} = g^{″}

l e t h (x) = g^{'} (x)

g^{″} = \frac{d h (x)}{d x} = \frac{d h (x)}{d g} \times \frac{d g}{d x} = h \frac{d h}{d g}

\Rightarrow g h = h \frac{d h}{d g}

i f h \neq 0 : (h = 0 \Rightarrow g (x) = c_{1} \Rightarrow y = c_{1} x + c_{2})

\Rightarrow g d g = d h

\Rightarrow \int g d g = \int d h

\Rightarrow \frac{g^{2} + c_{1}}{2} = h = \frac{d g}{d x}

\Rightarrow (g^{2} + c_{1}) d x = 2 d g

i f g^{2} + c_{1} \neq 0 :

\Rightarrow \frac{d g}{g^{2} + c_{1}} = \frac{d x}{2}

\Rightarrow \int \frac{d g}{g^{2} + c_{1}} = \int \frac{d x}{2}

i f c_{1} = 0 :

\Rightarrow \int \frac{d g}{g^{2}} = \int \frac{d x}{2}

\Rightarrow - \frac{1}{g} = \frac{x + c_{2}}{2}

\Rightarrow g = y^{'} = \frac{d y}{d x} = - \frac{2}{x + c_{2}}

\Rightarrow d y = - \frac{2 d x}{x + c_{2}}

\Rightarrow y = - 2 \ln ∣ x + c_{2} ∣ + c_{3}

i f c_{1} > 0, s a y c_{1} = c_{4}^{2}

\Rightarrow \int \frac{d g}{g^{2} + c_{4}^{2}} = \int \frac{d x}{2}

\Rightarrow \frac{1}{c_{4}} \tan^{- 1} \frac{g}{c_{4}} = \frac{x + c_{5}}{2}

\Rightarrow g = y^{'} = \frac{d y}{d x} = c_{4} \tan \frac{c_{4} (x + c_{5})}{2}

\Rightarrow y = \int c_{4} \tan \frac{c_{4} (x + c_{5})}{2} d x

\Rightarrow y = 2 \int \tan \frac{c_{4} (x + c_{5})}{2} d (\frac{c_{4} (x + c_{5})}{2})

\Rightarrow y = - 2 \ln ∣ \cos \frac{c_{4} (x + c_{5})}{2} ∣ + c_{6}

i f c_{1} < 0, s a y c_{1} = - c_{4}^{2}

\Rightarrow \int \frac{d g}{g^{2} - c_{4}^{2}} = \int \frac{d x}{2}

\Rightarrow \frac{1}{2 c_{4}} \ln ∣ \frac{g - c_{4}}{g + c_{4}} ∣= \frac{x + c_{5}}{2}

\Rightarrow \ln ∣ 1 - \frac{2 c_{4}}{g + c_{4}} ∣= c_{4} (x + c_{5})

\Rightarrow 1 - \frac{2 c_{4}}{g + c_{4}} = e^{c_{4} (x + c_{5})}

\Rightarrow \frac{2 c_{4}}{g + c_{4}} = 1 - e^{c_{4} (x + c_{5})}

\Rightarrow g = y^{'} = \frac{d y}{d x} = c_{4} [\frac{2}{1 - e^{c_{4} (x + c_{5})}} - 1]

\Rightarrow y = \int c_{4} [\frac{2}{1 - e^{c_{4} (x + c_{5})}} - 1] d x

\Rightarrow y = c_{4} [\int \frac{2 d x}{1 - e^{c_{4} (x + c_{5})}} - x]

\Rightarrow y = c_{4} [2 (x + c_{5}) - \frac{2 \ln ∣ 1 - e^{c_{4} (x + c_{5})} ∣}{c_{4}} - x]

\Rightarrow y = c_{4} x - 2 \ln ∣ 1 - e^{c_{4} (x + c_{5})} ∣ + c_{6}

i f g^{2} + c_{1} = 0 :

i f c_{1} = 0 :

\Rightarrow g = y^{'} = \frac{d y}{d x} = 0

\Rightarrow y = c_{2}

i f c_{1} > 0 : g^{2} + c_{1} \neq 0

i f c_{1} < 0 : g^{2} + c_{1} = 0

\Rightarrow g^{2} = - c_{1} = c_{2}^{2} s a y

\Rightarrow g = \pm c_{2}

\Rightarrow g = \frac{d y}{d x} = \pm c_{2}

\Rightarrow y = \pm c_{2} x + c_{3}

s u m m a r y :

f o l l o w i n g f u n c t i o n s a r e p o s s i b l e s o l u t i o n s :

y = f (x) = c_{1} x + c_{2}

y = f (x) = - 2 \ln ∣ x + c_{1} ∣ + c_{2}

y = f (x) = - 2 \ln ∣ \cos (c_{1} x + c_{2}) ∣ + c_{3}

y = f (x) = - 2 \ln ∣ 1 - e^{c_{1} x + c_{2}} ∣ + c_{1} x + c_{3}

Commented by MJS last updated on 02/Mar/19

great!

great!

Commented by otchereabdullai@gmail.com last updated on 02/Mar/19

Ideal Prof

Ideal Prof

Commented by malwaan last updated on 03/Mar/19

fantastic proof

fantastic proof

Commented by rahul 19 last updated on 03/Mar/19

W o n d e r f u l!

W o n d e r f u l!