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:

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

Answers (1)

Given, in the question

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

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

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

Now,

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

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

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

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

Now comparing the coefficients

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

Solving these:

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

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

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

Therefore,

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

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

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

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

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

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

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

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:
$y = \frac{1}{4} [\log {(x + 1)^{2} (x^{2} + 1)}] - \frac{1}{2} \tan^{- 1} x + 1$

Posted by

HARSH KANKARIA

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11. Find a particular solution satisfying the given condition: (x3+x2+x+1)dydx=2x2+x;y=1 when x=0

Answers (1)

Posted by

HARSH KANKARIA

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11. Find a particular solution satisfying the given condition:

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