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Form the differential equuations by eliminating arbitary constants given in brackets against each. $\large \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$=1 [a , b ] $

1 Answer

  • If we have an equation $f(x,y,c_1,c_2,....c_n)=u$ Containing n arbitrary constant $c_1,c_2...c_n$, then by differentiating n times, we get $(n+1)$ equations in total. If we eliminate the arbitrary constants $c_1,c_2....c_n,$ we get a D.E of order n
Step 1:
$\large\frac{2}{a^2}+\large\frac{2}{b^2}\bigg(\large\frac{dy}{dx}\bigg)^2+\frac{2y}{b^2}\frac{d^2 y}{dx^2}$$=0$-----(iii)
Step 2:
Eliminating $\large\frac{1}{a^2},\frac{1}{b^2}$ from (i),(ii),(iii)
$\begin{bmatrix}x^2 & y^2 & 1 \\ 2x & 2y \frac{dy}{dx} & 0 \\ 2 & 2 \bigg(\large\frac{dy}{dx}\bigg)^2 \normalsize +2y\frac{d^2y}{dx^2} & 0 \end{bmatrix}=0$
$4x \bigg[\bigg(\large\frac{dy}{dx}\bigg)^2 $$+y. \large\frac{d^2y}{dx^2}\bigg]-$$4y. \large\frac{dy}{dx}=0$
$xy \large\frac{d^2y}{dx^2}+x \bigg(\large\frac{dy}{dx}\bigg)^2$$-y \large\frac{dy}{dx}$$=0$ is the required D.E
answered Sep 5, 2013 by meena.p

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