dy/dx=1+x+y+xy y(-1)=0 then what is the value of funtion y ?

Question

AKASH GOYAL AskiitiansExpert-IITD · Answer

Dear Nikhil

Factorise RHS

RHS=1+x+y(1+x)= (1+x)(1+y)

dy/1+y = (1+x)dx

now integrate both sides

ln(1+y)=x+x²/2 + C

put x=-1 and y=0

0=-1 + 1/2 +C

C=1/2

ln(1+y)=x+x²/2 +1/2

1+y = e^{x+x2/2 +1/2}

y=e^{x+x2/2 +1/2}-1

All the best.

AKASH GOYAL

AskiitiansExpert-IIT Delhi

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vikas askiitian expert · Answer

dy/dx = 1+x+y+xy

dy/dx = 1+x +y(1+x)

dy/dx = (1+x)(1+y)

now using variable separation

dy/(1+y) = (1+x)dx

log(1+y) = (x+x²/2)+c

y(-1) = 0 so

c = 1/2

log(1+y) = x+x²/2+1/2

this is the required solution

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dy/dx=1+x+y+xy y(-1)=0 then what is the value of funtion y ?

dy/dx=1+x+y+xy
y(-1)=0
then what is the value of funtion y ?

2 Answers

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