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
0 votes
471 views
in Mathematics by (90.5k points)
closed by
Solve the differential equation. `(dy)/(dx) - y = e^(x)` Hence find the particluar solution for x = 0 and y = 1 .

1 Answer

0 votes
by (92.4k points)
selected by
 
Best answer
Given equation is `(dy)/(dx) - y= e^(x)`
The equation is of the type
`(dy)/(dx) +Py= Q`
where `" "P = - 1 and Q = e^(x)`
`therefore" "I.F = e^(int - 1dx) dx = e^(-x) = (1)/(e^(x))`
Multiplying both side of the differential equation by I.F., we get
`(1)/(e^(x)).[(dy)/(dx) - y] = (1)/(e^(x)).e^(x)`
`(1)/(e^(x)) (dy)/(dx) - (y)/(e^(x)) = 1`
`(d)/(dx) ((y)/(e^(x))) = 1`
Integrating both sides w.r.t.x, we get
`(y)/(e^(x)) = int 1 dx + c`
`(y)/(e^(x)) = x + c`
`ye^(-x) = x + c" "........(1)`
Which is the general soultion of the given equation.
Putting x = 0 and y = 1 in equations (1) , we get
`1 = 0 + c`
` c = 1`
Putting c = 1 in equations (1) , we get
`ye^(-x) = x + 1`
which is the particular solution.

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

...