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Form the differential equuations by eliminating arbitary constants given in brackets against each $y=(A+Bx)^{e^{\Large 3x}} [A , B ]$

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• 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:
$y=(A+Bx)e^{ 3x}$-----(i)
$\large\frac{dy}{dx}$$=3e^{3x} (A+Bx)+Be^{3x}-----(ii) =>\large\frac{dy}{dx}$$=3y+Be^{3x}$-----(iii)
=>$Be^{3x}=\large\frac{dy}{dx}$$-3y Step 2: Now from (iii) \large\frac{d^2y}{dx^2}$$=3 \large\frac{dy}{dx}$$+3 Be^{3x} \therefore \large\frac{d^2y}{dx^2}$$= 3 \large\frac{dy}{dx} $$+ 3 \bigg[\large\frac{dy}{dx}$$-3y\bigg]$
$\therefore \large\frac{d^2y}{dx^2} $$-6 \large\frac{dy}{dx}$$+9y=0$ is the required D.E
answered Sep 3, 2013 by