Reduce to variable separable form and solve: $\dfrac{dy}{dx}=\cos(x+y)$

Reduce to variable separable form and solve: (dy)/(dx)=cos(x+y)

Exercise 6.3 · Q49

\dfrac{dy}{dx}=\cos(x+y)

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Put $u=x+y$ , so $\dfrac{du}{dx}=1+\dfrac{dy}{dx}$ , i.e. $\dfrac{dy}{dx}=\dfrac{du}{dx}-1$ . The given equation is $\dfrac{dy}{dx}=\cos(x+y)$ , which becomes $\dfrac{du}{dx}-1=\cos u$ , i.e. $\dfrac{du}{dx}=1+\cos u=2\cos^2\!\left(\dfrac{u}{2}\right)$ . Separating: $\dfrac{1}{2}\sec^2\!\left(\dfrac{u}{2}\right)du=dx$ . Integrat …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL2 marks
$1+\dfrac{dy}{dx}=\csc(x+y)$
›Reveal solutionSolution

Put $u=x+y$ ; the equation becomes $\dfrac{du}{dx}=\csc u$ , separable.

$u=x+y \Rightarrow \dfrac{du}{dx}=1+\dfrac{dy}{dx}=\csc u$

$\sin u\,du=dx$

…

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CBSE 2023Set ANNUAL2 marksMCQ
$\dfrac{dx}{dt} = \dfrac{x\log x}{t}$
(a) $x = e^{ct}$
(b) $x + e^{ct} = 0$
(c) $x = e^t + t$
(d) $xe^{ct} = 0$
›Reveal solutionSolution

Separate variables and integrate.

$\dfrac{dx}{x\log x}=\dfrac{dt}{t}$

Integrating: $\log(\log x)=\log t+\log k=\log(kt)$

…

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CBSE 2023Set ANNUAL2 marks
$\cos x \cos y\,dy - \sin x \sin y\,dx = 0$
›Reveal solutionSolution

Separate variables: $\cot y\,dy=\tan x\,dx$ .

$\cos x\cos y\,dy=\sin x\sin y\,dx \Rightarrow \cot y\,dy=\tan x\,dx$

Integrating: $\log|\sin y|=-\log|\cos x|+\log k$

…

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CBSE 2022Set ANNUAL2 marks
$y\dfrac{dy}{dx} + x = 0$
›Reveal solutionSolution

Separate variables and integrate.

$y\dfrac{dy}{dx}+x=0 \implies y\,dy = -x\,dx$

…

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CBSE 2020Set ANNUAL2 marks
$\dfrac{dy}{dx} = x^2y + y$
›Reveal solutionSolution

Separate variables and integrate.

$\dfrac{dy}{dx} = y(x^2+1)$

Separating variables:

$\dfrac{dy}{y} = (x^2+1)\,dx$

Integrating both sides: …

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←Previous question · 6.3Q48 — Find the particular solution: cos((dy)/(dx))=a,a∈[1,2],y(0)=2.→Next question · 6.3Q50 — Reduce to variable separable form and solve: (x-y)²(dy)/(dx)=a²

Q.Reduce to variable separable form and solve: $\dfrac{dy}{dx}=\cos(x+y)$

Concept understanding — Differential Equations with Variables Separable

Q.Reduce to variable separable form and solve: dydx=cos⁡(x+y)\dfrac{dy}{dx}=\cos(x+y)dxdy​=cos(x+y)

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Q.Reduce to variable separable form and solve: $\dfrac{dy}{dx}=\cos(x+y)$

Unlock everything free for 14 days