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Find a particular solution satisfying the given condition $x(x^2-1)\large\frac{dy}{dx}$$=1;y=0\;when\;x=2$

$\begin{array}{1 1} y=\large\frac{ 1}{2}\log\large\frac{(x^2-1)}{x^2} - \frac{1}{2}\log\large\frac{3}{4} \\y=\large\frac{ 1}{4}\log\large\frac{(x^2+1)}{x^2} - \frac{1}{2}\log\large\frac{3}{4} \\y=\large\frac{ 1}{2}\log\large\frac{(x^2-1)}{x^2} + \frac{1}{2}\log\large\frac{3}{4} \\ y=\large\frac{ 1}{2}\log\large\frac{(x^2-1)}{x^2} - \frac{1}{4}\log\large\frac{3}{4} \end{array} $

1 Answer

Toolbox:
  • Rational function is defined as the ratio of two polynomials in the form of $\large\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials in x. If the degree of p(x) is less than degree of q(x) then it is said to be proper. Such fractions can be evaluated by breaking in factors given as $(x+a) (x+b)(x+c)$ as $\large\frac{A}{(x+a)}$$ +\large\frac{ B}{(x+b)}$$ +\large\frac{ C}{(x+c)}$
Step 1:
Given: $x(x^2 - 1)\large\frac{dy}{dx }$$= 1; y = 0\; when\; x = 2$
Seperating the variables we get,
$dy= \large\frac{ dx}{(x)(x^2 - 1)}$
$x^2 - 1$ can be factorised as $(x+1)(x-1)$
$dy =\large\frac{ dx}{x(x-1)(x+1)}$
Step 2:
Using the information from the tool box we get
$\large\frac{1}{x(x-1)(x+1)} =\frac{ A}{x} +\frac{ B}{(x-1) } +\frac{ C}{x+1}$
$1 = A(x-1)(x+1) + Bx(x-1) + C x(x+1)$
$x^2 = (A+B+C) x^2$
$x = x(-B + C)$
$1 = -$A
Step 3:
comparing the coefficients $1 = A+B+C$
$1 = -B + C$
$1 = -A$
On solving these equations we get $A = -1, B = \large\frac{1}{2}$ and $C = \large\frac{1}{2}$
Substituting these values we get
$dy = -\large\frac{dx}{x} +\frac{ dx}{2(x+1)} +\frac{dx}{2(x-1)}$
Step 4:
Integrating on both sides
$y = -\log x + (1/2)\log(x+1) + (1/2)\log(x-1) + $C
$y =\large\frac{ 1}{2}$$\log\frac{ (x+1)(x-1)}{x^2}$
To find the value of C let us substitute the value of $x = 2$ and $y = 0$
$0 = -1/2\log\large\frac{ (4 -1)}{4}$
$C= \large\frac{-1}{2}$$\log\large\frac{3}{4}$
Hence the required solution is $y=\large\frac{ 1}{2}$$\log\large\frac{(x^2-1)}{x^2} - \frac{1}{2}$$\log\large\frac{3}{4}$
answered Aug 14, 2013 by sreemathi.v
 
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