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Exercise 6.1 · Q10

Q.Determine the order and degree of the differential equation: x+d2ydx2=1+(d2ydx2)2x+\dfrac{d^2y}{dx^2}=\sqrt{1+\left(\dfrac{d^2y}{dx^2}\right)^2}

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The equation is x+d2ydx2=1+(d2ydx2)2x+\dfrac{d^2y}{dx^2}=\sqrt{1+\left(\dfrac{d^2y}{dx^2}\right)^2}. Squaring both sides: (x+d2ydx2)2=1+(d2ydx2)2\left(x+\dfrac{d^2y}{dx^2}\right)^2=1+\left(\dfrac{d^2y}{dx^2}\right)^2, i.e. x2+2xd2ydx2+(d2ydx2)2=1+(d2ydx2)2x^2+2x\dfrac{d^2y}{dx^2}+\left(\dfrac{d^2y}{dx^2}\right)^2=1+\left(\dfrac{d^2y}{dx^2}\right)^2. The squared terms in d2ydx2\dfrac{d^2y}{dx^2} cancel from both sides, leaving 2xd2ydx2=1−x22x\dfrac{d^2y}{dx^2}=1-x^2 — a genuine simplification, not a trick to avoid. …

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