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Exercise 6.1 · Q4

Q.Determine the order and degree of the differential equation: d2ydx2+dydx+x=1+d3ydx3\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}+x=\sqrt{1+\dfrac{d^3y}{dx^3}}

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The equation is d2ydx2+dydx+x=1+d3ydx3\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}+x=\sqrt{1+\dfrac{d^3y}{dx^3}}. Squaring both sides removes the square root: (d2ydx2+dydx+x)2=1+d3ydx3\left(\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}+x\right)^2=1+\dfrac{d^3y}{dx^3}. The highest derivative present is d3ydx3\dfrac{d^3y}{dx^3}, giving order 33; it …

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