Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find a particular solution satisfying the given condition: (x^3 + x^2 + x + 1)dy/dx = 2x^2 + x; y = 1 when x = 0

11.  Find a particular solution satisfying the given condition:

$\left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x ; y=1$ when $x=0$

Answers (1)

best_answer

Given, in the question

$\left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x$

$\Longrightarrow \int d y=\int \frac{2 x^2+x}{\left(x^3+x^2+x+1\right)} d x$

$\left(x^3+x^2+x+1\right)=(x+1)\left(x^2+1\right)$

Now,

$\Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+1}$

$\Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A x^2+A(B x+C)(x+1)}{(x+1)\left(x^2+1\right)}$

$\Rightarrow 2 x^2+x=A x^2+A+B x+C x+C$

$\Rightarrow 2 x^2+x=(A+B) x^2+(B+C) x+A+C$

Now comparing the coefficients

$A + B = 2;  B + C = 1;  A + C = 0$

Solving these:

$\mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{3}{2}, \mathrm{C}=-\frac{1}{2}$

Putting the values of $A,B,C$:

$\Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{1}{2} \frac{1}{(x+1)}+\frac{1}{2} \frac{3 x-1}{x^2+1}$

Therefore,

$\Rightarrow \int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^2+1} d x$

$\Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{2} \int \frac{\mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}-\frac{1}{2} \int \frac{\mathrm{dx}}{\mathrm{x}^2+1}$

$\Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{4} \int \frac{2 \mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}-\frac{1}{2} \tan ^{-1} \mathrm{x}$

Let $\mathrm{x}^2+1=\mathrm{t}$

$\therefore \frac{3}{4} \int \frac{2 \mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}=\frac{3}{4} \int \frac{\mathrm{dt}}{\mathrm{t}}$

So, $\mathrm{I}=\frac{3}{4} \log \mathrm{t}$

$\mathrm{I}=\frac{3}{4} \log \left(\mathrm{x}^2+1\right)$
$\Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{4} \log \left(\mathrm{x}^2+1\right)-\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c}$
$\Rightarrow \mathrm{y}=\frac{1}{4}\left[2 \log (\mathrm{x}+1)+3 \log \left(\mathrm{x}^2+1\right)\right]-\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c}$
$\Rightarrow \mathrm{y}=\frac{1}{4}\left[\log (\mathrm{x}+1)^2+\log \left(\mathrm{x}^2+1\right)^3\right]-\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c}$

Now, $y= 1$ when $x = 0$

$1=\frac{1}{4} \times 0-\frac{1}{2} \times 0+c$
$c=1$
Putting the value of $c$ , we get:
$\mathrm{y}=\frac{1}{4}\left[\log \left\{(\mathrm{x}+1)^2\left(\mathrm{x}^2+1\right)\right\}\right]-\frac{1}{2} \tan ^{-1} \mathrm{x}+1$

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question