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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find a particular solution satisfying the given condition $(x^3+x^2+x+1)\large\frac{dy}{dx}$$=2x^2+x;y=1\;when\;x=0$

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1 Answer

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Toolbox:
  • Rational functions 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 $q(x)$, then it is called proper fraction.
  • Such fractions can be evaluated by breaking in factors given as follows: $px +\large\frac{ q}{(x+a)(x^2+1) }=\large\frac{ A}{(x+a)} +\frac{ (Bx+C)}{(x^2 +1) }$
Step 1:
Given: $[ x^3 + x^2 + x + 1] \large\frac{dy}{dx} =$$ 2x^2 +x$; when $y = $1 and x = 0
$x^3 + x^2 + x + $1 can be factorised as $(x+1)(x^2 + $1)
on rearranging we get,$\large\frac{ dy}{dx} =\frac{ [2x^2 + x]}{(x^2 + 1) (x + 1) }$
on seperating the variables we get,
$dy =\large\frac{ [2x^2 + x) dx}{(x^2 + 1)(x+1)}$
Step 2:
On integrating we get,
$\int dy=\int\large\frac{ (2x^2 + x)dx}{(x^2 + 1)(x + 1) }$
Using the information in the tool box, we can integrate the above function on the RHS by the method of partial fractions
$\large\frac{(2x^2 +x)}{(x^2 + 1)(x +1)(x^2 + 1)} = \large\frac{A}{(x+1)} +\frac{ [Bx+C]}{(x^2 + 1) }$
$\Rightarrow A(x^2 + 1) + (Bx + C)(x+1)$
$2x^2 + x = x^2(A+B) + x(A+C) + A+C$
Step 3:
Comparing the coefficients of $x^2$,
$2 = A+B$
Comparing the coefficients of x,
$1 = A+B$
Comparing the coefficients of the constant term,
$A+B = 0$
On solving the three linear equations we get the value of $A =\large\frac{ 1}{2}$, $B= \large\frac{3}{2}$, and $C =\frac{ -1}{2}$.
Now substituting the values of A,B,C we get,
$y = \int\large\frac{ 1}{2(x+1)} +\frac{ 3x}{2 }-\frac{ 1}{2/[x^2 + 1]}dx$
$\Rightarrow\large\frac{ 1}{2}\int\large\frac{ dx}{(x+1) }+\frac{ 3}{2}$$\int xdx/x^2 + 1 - 1/2 \int\large\frac{ dx}{x^2 + 1}$
y = $\large\frac{1}{2}$$ \log|x + 1| + (3/4)\log|x^2 + 1| - 1/2 \tan^-1x + C$
Step 5:
Now to find the value of C let us substitute the value of x and y we get,
$1 = (1/2)\log|0+1| + 3/2\log|0+ 1| - 1/2 \tan^{-1} + C$ ( because $ \log 1 = 0 $and $\tan^{-1} = 0)$
C = 1
answered Aug 15, 2013 by sreemathi.v
edited Aug 15, 2013 by sreemathi.v
 
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