Q.13 Solve : 1 + dy/dx = cosec(x + y) ; put x + y = u .

Answer: x + cos(x + y) = c Put x + y = u . Then 1 + dy/dx = du/dx , so the equation becomes du/dx = cosec u , i.e. sin u du = dx . Integrating: ∫ sin u du = ∫ dx ⇒ -cos u = x + c₁ . Resubstituting u = x + y : x + cos(x + y) = c , where c = -c₁ . This is the general solution.

Solve : \(1 + \frac{dy}{dx} = \operatorname{cosec}(x + y)\) ; put \(x + y = u\).

Q.13 Solve : \(1 + \frac{dy}{dx} = \operatorname{cosec}(x + y)\) ; put \(x + y = u\).

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

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Answer: \(x + \cos(x + y) = c\)

Put \(x + y = u\). Then \(1 + \frac{dy}{dx} = \frac{du}{dx}\), so the equation becomes \(\frac{du}{dx} = \operatorname{cosec} u\), i.e. \(\sin u\,du = dx\).

Integrating: \(\int \sin u\,du = \int dx \Rightarrow -\cos u = x + c_1\).

Resubstituting \(u = x + y\): \(x + \cos(x + y) = c\), where \(c = -c_1\). This is the general solution.

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Solve : \(1 + \frac{dy}{dx} = \operatorname{cosec}(x + y)\) ; put \(x + y = u\).

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