d-2-y-dx-2-log-y-0-

Question Number 115298 by Dwaipayan Shikari last updated on 24/Sep/20

\frac{d^{2} y}{{dx}^{2}} + \log (y) = 0

Commented by mohammad17 last updated on 25/Sep/20

y^{^{″}} + l n y = 0 \Rightarrow y^{^{″}} = - l n y

\int d (y^{^{'}}) = - \int l n y d y

y^{^{'}} = - y l n y + y + C_{1}

\int d (y) = \int (C_{1} + y - y l n y) d y

y = C_{1} y + \frac{1}{2} y^{2} - \frac{1}{2} y^{2} l n y + \frac{1}{4} y^{2} + C_{2}

y = \frac{3}{4} y^{2} - \frac{1}{2} y^{2} l n y + C_{1} y + C_{2}

y = y^{2} (\frac{3 - 2 l n y}{4}) + C_{1} y + C_{2}

⟨ m . t ⟩

Commented by Olaf last updated on 26/Sep/20

s o r r y m i s t e r .

\int d (y^{'}) = - \int l n y d x \dots .!

a n d n o t - \int l n y d y

A n d a s y o u c a n s e e,

i f y o u d e r i v a t e y o u r f i n a l r e s u l t

t o v e r i f y t h e i n i t i a l e q u a t i o n

u n a f o r n a t e l y {i t}^{'} s w r o n g .

Answered by Olaf last updated on 24/Sep/20

y^{″} + \ln y = 0

y^{″} y^{'} + y^{'} \ln y = 0

\int y^{″} y^{'} d x + \int y^{'} \ln y d x = 0

\frac{1}{2} y^{' 2} + y \ln y - \int y \frac{y^{'}}{y} = C_{1}

\frac{1}{2} y^{' 2} + y \ln y - y = C_{1}

y^{' 2} = 2 y (1 - \ln y) + C_{2}

\frac{y^{'}}{\sqrt{2 y (1 - \ln y) + C_{2}}} = 1

\int \frac{y^{'}}{\sqrt{2 y (1 - \ln y) + C_{2}}} d x = x + C_{3}

Commented by Dwaipayan Shikari last updated on 25/Sep/20

Is there any definite solution ?

Commented by Olaf last updated on 26/Sep/20

n o s i r .

o n l y a p p r o x i m a t e s o l u t i o n s

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