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Question 163 of 177

Q.Solve x+ydydx=sec⁡(x2+y2)x + y\dfrac{dy}{dx} = \sec(x^2+y^2)

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022Subjective· 3mImportance★★★★★
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Substitute v=x2+y2v=x^2+y^2 to reduce to a separable equation.

Let v=x2+y2v=x^2+y^2, so dvdx=2x+2ydydx=2(x+ydydx)\dfrac{dv}{dx}=2x+2y\dfrac{dy}{dx} = 2\left(x+y\dfrac{dy}{dx}\right).

Given x+ydydx=sec⁡vx+y\dfrac{dy}{dx}=\sec v:

dvdx=2sec⁡v\dfrac{dv}{dx} = 2\sec v

Separating variables: …

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