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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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For the differential equation $xy\large\frac{dy}{dx}=$$(x+2)(y+2),$ find the solution curve passing through the point $(1,-1)$.

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

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1 Answer

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Toolbox:
  • Whenever a function occurs in the form $\int\large\frac{x}{x+a}$, then we can integrate by adding and subtracting a to the numerator.
Step 1:
given $xy\large\frac{dy}{dx} $$= (x+2)(y+2) $
On rearranging we get $\large\frac{ydy}{y+2} =\frac{ (x+2)dx}{x}$
Using the information in the tool box,
$\large\frac{(y+2-2)dy}{y+2 }= $$dx + \large\frac{2dx}{x}$
$\large\frac{(y+2)dy}{y+2 }-\frac{ 2dy}{y+2}$$ = dx + \large\frac{2dx}{x}$
Step 2:
On integrating both sides we get,
$\int dy - \int \large\frac{2dy}{y+2 }$$= \int dx +\int\large\frac{2dx}{x }$
$y -2\log(y+2) = x + 2\log x + C$
But $2 \log x =\log(x^2)$ and $2\log(y+2) =\log(y+2)^2$
$y -x + C = log(x^2)(y+2)^2$
Step 3:
Now substituting the value of $x = $1 and $y = -$1 to find the value of $C$
$-1 - 2\log(-1+2) = 1 + 2\log 1 + C$
Since $\log 1 = 0$, we get the value of $C = -$2
on substituting the value for $C = -2$ we get,
$y - x +2 = log(x^2)(y+2)^2$
This is the required general solution.
answered Aug 14, 2013 by sreemathi.v
 

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