yy′′−(y′)^2 =e^(ax ) find genral solution?


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Question Number 126896 by BHOOPENDRA last updated on 25/Dec/20

${y y}^{″} - {(y^{'})}^{2} = e^{a x} f i n d g e n r a l s o l u t i o n ?$
Commented by liberty last updated on 25/Dec/20

$o k i t r y s o l v e i t .$ $f o r e q u a t i o n {y y}^{″} - {(y^{'})}^{2} = e^{a x}$ $o r y^{″} - \frac{{(y^{'})}^{2}}{y} = \frac{e^{a x}}{y}$ $p u t y^{'} = v (y); w h e r e y = f (x)$ $t h e n y^{″} = v (y) v^{'} (y) . t h e e q u a t i o n b e c o m e s$ $v^{'} - \frac{v}{y} = \frac{e^{a x}}{y} . t h e t e r m e^{a x} c a n b e c o n s i d e r e d$ $a s a c o n s t a n t . T h e i n t e g r a t i n g f a c t o r i s \frac{1}{y}$ $a n d t h e s o l u t i o n g i v e n b y$ $v = y [\int \frac{e^{a x}}{y^{2}} d y + b]$ $v = b y - e^{a x}; g i v e \frac{d y}{d x} - b y = - e^{a x}$ $f o r t h i s e q u a t i o n i n t e g r a t i n g f a c t o r$ $i s e^{- b x} a n d t h e s o l u t i o n i s$ $y = e^{b x} [\int (e^{- a x}) (e^{- b x}) d x + c]$ $y = \frac{e^{a x}}{b - a} + c . e^{b x}, b \neq a$
Commented by mr W last updated on 25/Dec/20

$h o w d i d y o u g e t$ $v^{'} - \frac{v}{y} = \frac{e^{a x}}{y}$ $f r o m$ $y^{″} - \frac{{(y^{'})}^{2}}{y} = \frac{e^{a x}}{y} ?$ $i t s h o u l d b e$ ${v v}^{'} - \frac{v^{2}}{y} = \frac{e^{a x}}{y}$ $i {d o n}^{'} t t h i n k y o u r s o l u t i o n i s c o r r e c t .$
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