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Form the differential equuations by eliminating arbitary constants given in brackets against each. $y=ax^{2} +bx + c[a , b ]$

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:
$y=ax^2 +bx+c$-----(i)
Where a,b are arbitrary constants
$\large\frac{dy}{dx}$$=2ax+b $ -----(ii)
Step 2:
from (ii) and (iii)
$\qquad= \large\frac{1}{2} \frac{d^2 y}{dx^2}$
and $\large\frac{dy}{dx}=\frac{d^2 y}{dx^2}$$.x+b$
=>$b= \large\frac{dy}{dx}$$- x \large\frac{d^2 y}{dx^2}$
Step 3:
Substitute in (i) for a,b
$y= \large\frac{x^2}{2} \frac{d^2y}{dx^2}$$+x\bigg(\large\frac{dy}{dx}-x \frac{d^2y}{dx^2}\bigg)$$+c$
$\therefore \large\frac{d^2y}{dx^2}\bigg[ \frac{x^2}{2} $$-x^2\bigg]+x \large\frac{dy}{dx}$$-y +c=0$
$\therefore \large\frac{x^2}{2} \frac{d^2 y}{dx^2}-x \frac{dy}{dx}$$+y-c=0$ is the required D.E


answered Sep 3, 2013 by meena.p

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