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Q.Find the order and degree of differential equation y = dy/dx + sqrt(1 + (dy/dx)^2).

Chhattisgarh CgbseCGBSE Intermediate Board 2024Subjective· 1mImportance★★★★★
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Isolate the radical, square both sides, and simplify — the squared derivative term cancels, leaving a first-degree equation.

Let y′=dy/dxy' = dy/dx. The equation is y=y′+1+(y′)2y = y' + \sqrt{1+(y')^2}.

Isolate the radical: y−y′=1+(y′)2y - y' = \sqrt{1+(y')^2}

Square both sides (to remove the radical, giving a polynomial equation in derivatives):

(y−y′)2=1+(y′)2(y-y')^2 = 1+(y')^2

y2−2yy′+(y′)2=1+(y′)2y^2 - 2yy' + (y')^2 = 1 + (y')^2

The (y′)2(y')^2 terms cancel from both sides, leaving:

y2−2yy′−1=0y^2 - 2yy' - 1 = 0

This is a polynomial equation in y′y' (the highest, i.e. first-order, derivative present) in which y′y' occurs to the power 1.

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