The general solution of the differintial equation : xdy/dx-ay=1(a≠0) is Option 1: y=-1/a+Cx^a:Where C is an arbitrary constant Option 2: y=-1/a+Cx^-a:Where C is an arbitrary constant Option 3: y=-1/a+Cx:Where C is an arbitrary constant Option 4: y=-1/a+Cx^-a-1:Where C is an arbitrary constant

The answer is 'y=-1/a+Cx^a:Where C is an arbitrary constant'. View full solution:

CUET Mathematics - The general solution of the differintial equation : xdy/dx-ay=1(a≠0) is | PYQs + Solutions

CUET Mathematics 2026 14th May Shift 2

Calculus

Medium

core

$y=-\frac{1}{a}+Cx^a$ :Where C is an arbitrary constant

$y=-\frac{1}{a}+Cx^{-a}$ :Where C is an arbitrary constant

$y=-\frac{1}{a}+Cx$ :Where C is an arbitrary constant

$y=-\frac{1}{a}+Cx^{-a-1}$ :Where C is an arbitrary constant

✅ Correct Option: 1

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The solution of the differential equation $x\frac{dy}{dx} = 2x^2 + 1$ ; $x \neq 0$ with $y = 1$ when $x = 1$ , is

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Solution of $\frac{dy}{dx} = (1+x^2)(1+y^2)$ is:

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Match List-I with List-II

List-I	List-II
(A) $y + \frac{dy}{dx} = \frac{1}{4}\int y\,dx$	(I) Order 1, Degree 1
(B) $y = \frac{dy}{dx} + \frac{C}{dy/dx}$	(II) Order 2, Degree 2
(C) $(xy^2+x)dx + (y-x^2)dy = 0$	(III) Order 2, Degree 1
(D) $\left[\left\{1+\left(\frac{dy}{dx}\right)^2\right\}^{3/2}/\frac{d^2y}{dx^2}\right] = K$	(IV) Order 1, Degree 2

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