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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The solution of the differential equation $\Large \frac{dy}{dx}=\frac{1-y^2}{1-x^2}$ is:\begin{array}{1 1}(A)\;y=\tan^{-1}x & (B)\;y-x=k(1+xy)\\(C)\;x=tan^{-1}y & (D)\;\tan (xy)=k\end{array}

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Toolbox:
  • If a linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x),$ then ot can be solved by seperating the variables and then integrating.
  • $\int \large\frac{dx}{x+a}$$=\log |x+a|$
Given $ \large\frac{dy}{dx}=\frac{1-y^2}{1-x^2}$
Seperating the variables we get
$\large\frac{dy}{(1-y^2)}=\frac{dx}{1-x^2}$
$\int \large\frac{dx}{a^2-x^2}=\frac{1}{2a} $$\log \large\frac{a+x}{a-x}$
integrating on both sides,
$\int \large\frac{dy}{(1-y^2)}=\int \frac{dx}{1-x^2}$
$\large\frac{1}{2} $$\log \large\bigg|\frac{1+y}{1-y}\bigg|=\frac{1}{2} $$\log \large\bigg|\frac{1+x}{1-x}\bigg|$$+\log c$
$=>\log \large\bigg|\frac{1+y}{1-y}\bigg|= $$\log 2c \large\bigg|\frac{1+x}{1-x}\bigg|$
$=> \large\frac{(1+y)}{(1-y)}$$=2c \large\frac{(1+x)}{(1-x)}$$ \qquad (Let \;2c=k)$
$=> \large\frac{(1+y)}{(1-y)}=k\frac{(1+x)}{(1-x)}$
Cross mutiplying we get,
$(1+y)(1-x)=k(1+x)(1-y)$
$1-x+y-xy=k[1-y+x-xy]$
$-x+y-xy=(k-1)+k(x-y)-kxy \qquad $[(k-1) is a constant]
$=>y-x-xy=k(x-y)-kxy$
$=>y-x=k(x-y)+xy(k+1)$
But again $k+1$ is a constant
Hence $y-x=k(x-y)+xy$
Dividing throughout by $(x-y)$ we get
$-1=k+\large\frac{xy}{x-y}$
$=>-(x-y)=k(x-y)+xy$
$=>y-x=k(1+xy)$
Hence $B$ is the correct option
answered May 17, 2013 by meena.p
 

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