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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form the differential equation by eliminating A and B in $Ax^2+By^2=1$.

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Toolbox:
  • If the given equation has n arbitary constans, then its differential equation will be of $n^{th}$ order
  • $\large\frac{d}{dx}$$(uv)=\large\frac{du}{dx}$$.v+\large\frac{dv}{dx}$$.u$
Given $Ax^2+By^2=1$-----(1)
We are asked to find the differential equation by eliminating A and B
Consider $Ax^2+By^2=1$
Differentiate with respect to x
$2Ax+2By.y'=0$=>$Ax=-Byy'$-----(2)
Again differentiate with respect to x
$yy'$ is in the product form, hence apply product rule
$\large\frac{d}{dx}$$(uv)=\large\frac{d}{dv}$$(u).v+\large\frac{d}{dx}$$(v).u$
Here $u=y$ hence $u'=y'$
and $v=y'$ hence $v'=y''$
Therefore $\large\frac{d}{dx}$$(yy')=y.y''+y'.y'$
$=(y')^2+y.y''$
Therefore differentiating $Ax=-Byy'$ we get
$A=-B[(y')^2+y.y'']$-----(3)
Now substituting the value of A in equ(2)
$-B[(y')^2+y.y'']x=-Byy''$
=>$x(y')^2+y.y''x=yy''$
=> $x(y')^2+y.y''x-yy''=0$
or => $x(y')^2+y.y''(x-1)=0$

 

answered May 10, 2013 by meena.p
 

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