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Example · Example 2

Q.Find the order and degree (if defined) of the differential equation dydx+sin⁡ ⁣(dydx)=0\dfrac{dy}{dx} + \sin\!\left(\dfrac{dy}{dx}\right) = 0.

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✓ Free question

The highest (and only) derivative present is dydx\dfrac{dy}{dx}, so the order is 11.

To read off a degree, the equation must be expressible as a polynomial in the derivatives. Here dydx\dfrac{dy}{dx} appears both on its own and inside sin⁡ ⁣(dydx)\sin\!\left(\dfrac{dy}{dx}\right) -- and there is no algebraic rearrangement (no squaring, cubing, or clearing of denominators) that removes the sine and leaves a polynomial in dydx\dfrac{dy}{dx}, since sin⁡\sin is a transcendental function. Therefore the degree is not defined for this equation.

✓Final answer

Order =1= 1; degree is not defined.

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