Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

11. Find a particular solution satisfying the given condition:

$(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; \ y = 1\ \textup{when}\ x = 0$

Answers (1)

best_answer

Given, in the question

$(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x$

$\\ \implies \int dy = \int\frac{2x^2 + x}{(x^3 + x^2 + x + 1)}dx$

$(x^3 + x^2 + x + 1) = (x +1)(x^2+1)$

Now,

Now comparing the coefficients

A + B = 2; B + C = 1; A + C = 0

Solving these:

Putting the values of A,B,C:

Therefore,

Now, y= 1 when x = 0

c = 1

Putting the value of c, we get:

Posted by

HARSH KANKARIA

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th

anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question