wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The solution curve of the differential equation, (1+e-x)(1+y2)dydx=y2, which passes through the point (0,1), is


A

y2=1+ylog1+e-x2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

y2-1=ylog1+e-x2+2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

y2-1=ylog1+ex2+2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

y2=1+ylog1+ex2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D

y2=1+ylog1+ex2


Explanation for the correct option

Given: (1+e-x)(1+y2)dydx=y2

1+y2y2dy=11+e-xdx

integrating both sides

1+y2y2dy=11+e-xdx1y2+1dy=exex+1dx

Let ex+1=t

ex·dx=dt

1y2dx+1dy=1tdty-1-1+y=log(t)+c

Put t=1+ex

-1y+y=log(1+ex)+c

Since, the curve passes through the point (0,1)

Put y=1,x=0

-11+1=log(1+e0)+c0=log(2)+cc=-log(2)

therefore, the equation becomes

-1y+y=log(1+ex)-log(2)y2-1y=log1+ex2y2-1=ylog1+ex2

Hence, option D is correct.


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon