Reduce to variable separable form and solve: (2x-2y+3)dx-(x-y+1)dy=0, when x=0,y=1.

Exercise 6.3 · Q53

(2x-2y+3)dx-(x-y+1)dy=0

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Put $u=x-y$ , so $dy=dx-du$ . The equation $(2x-2y+3)dx-(x-y+1)dy=0$ becomes $(2u+3)dx-(u+1)(dx-du)=0$ , i.e. $(u+2)dx+(u+1)du=0$ , so $dx=-\dfrac{u+1}{u+2}du=-\left(1-\dfrac{1}{u+2}\right)du$ . Integrating: $x=-u+\log|u+2|+c$ , i.e. $x=-(x-y)+\log|x-y+2|+c$ , which simplifie …

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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 · 6.3Q52 — Reduce to variable separable form and solve: cos²(x-2y)=1-2(dy)/(dx)→Next section · 6.4Solution of Differential Equation

Q.Reduce to variable separable form and solve: $(2x-2y+3)dx-(x-y+1)dy=0$ , when $x=0,\ y=1$ .

Concept understanding — Differential Equations with Variables Separable

Q.Reduce to variable separable form and solve: (2x−2y+3)dx−(x−y+1)dy=0(2x-2y+3)dx-(x-y+1)dy=0(2x−2y+3)dx−(x−y+1)dy=0, when x=0, y=1x=0,\ y=1x=0, y=1.

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Q.Reduce to variable separable form and solve: $(2x-2y+3)dx-(x-y+1)dy=0$ , when $x=0,\ y=1$ .

Unlock everything free for 14 days