Answer :
Formula:-
(i) if a differential equation is ,
then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e∫Pdx
(ii) ∫dx = x + c
This is a linear differential equation, comparing it with
P = 1, Q = e–2x
I.F = e∫Pdx
= e∫Pdx
= ex
Solution of the equation is given by
y(I.F) = ∫Q.(I.F)dx + c
⇒ y(ex) = ∫e–2x ex dx + c
⇒ y(ex) = ∫e–x dx + c
⇒ y(ex) = –e–2x + c
⇒ y = –e–2x + c e–x
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
view all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation


RELATED QUESTIONS :
Integrating facto
Mathematics - ExemplarSolve the followi
RD Sharma - Volume 2Solve the followi
Mathematics - Board PapersSolve the followi
RD Sharma - Volume 2Solve the followi
RD Sharma - Volume 2Solve the followi
RD Sharma - Volume 2Solve the followi
RD Sharma - Volume 2Solve the followi
RD Sharma - Volume 2Solve the followi
Mathematics - Board PapersSolve the followi
RD Sharma - Volume 2