Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
84 views
in Mathematics by (88.8k points)
closed by
हल करें: `2ye^(x//y) dx + (y-2xe^(x//y))dy =0`, यदि `y=1` जब `x=0`

1 Answer

+1 vote
by (87.6k points)
selected by
 
Best answer
दिया गया अवकल समीकरण है: `2ye^(x//y) dx + (y-2xe^(x//y))dy=0`
या, `(dx)/(dy) =(2xe^(x/y)-y)/(2ye^(x//y)) = (2.(x/y)e^(x/y)-1)/(2e^(x/y))`............(1)
स्पष्तः (1) एक समघातीय अवकल समीकरण है|
`x=vy` रखें, तो `(dx)/(dy)=v+y(dv)/(dy)`
x तथा `(dx)/(dy)` का मान (1) एक समघातीय समीकरण है|
x तथा `(dx)/(dy)` का मान (1) में रखने पर हमें मिलता है,
`v+y(dv)(dy) =(2ve^(y)-1)/(2e^(v))`
`rArr y(dv)/(dx) = (2ve^(v)-1)/(2e^(v)) - v=-1/(2e^(v)) rArr 2e^(v)dv = -(dy)/y`
`rArr int 2e^(v)dv = -int (dv)/y rArr 2e^(v) = -log|y|+C`...........(2)
`rArr 2e^(x/y) + log|y|=C`
प्रशन से, जब `x=0, y=1`
`therefore (2)` से, `2e^(0) + log1 =C rArr 2.1 + 0 = C rArr C=2`
(2) में C के इस मान को रखने पर हमें मिलता है,
`2e^(x/y) + log|y| =2`, यही अभीष्ट हल है|

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...