New chat PYQ Library Doubts Rank report

Sign in to see Recents

Sign in to save progress →

Differential Equations question

2023 · 29 Jan · Shift 1 · Q32

Guest · try one question or challengeGuest · try one question or challenge

PYQ Library
JEE Main
Mathematics
Differential Equations
2023 · 29 Jan · Shift 1 · Q32

Differential Equations question

2023 · 29 Jan · Shift 1 · Q32

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

Let

y=f(x)

be the solution of the differential equation

y(x+1)dx-x^2dy=0,y(1)=e

. Then

\mathop {\lim }\limits_{x \to {0^ + }} f(x)

is equal to

A
${e^2}$
B
0
C
${1 \over {{e^2}}}$
D
${1 \over e}$

More from Differential Equations

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to2023 · MCQ
Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} - \frac{3 x ^{5} t a n ^{- 1} ( x ^{3} )}{( 1 + x ^{6} ) ^{3/2}} y = 2 x exp {\frac{x ^{3} - t a n ^{- 1} x ^{3}}{( 1 + x ^{6} )}} pass through the origin. Then y (1) is equal to :$ …2023 · MCQ
The solution of the differential equation $\frac{d y}{d x} = - (\frac{x ^{2} + 3 y ^{2}}{3 x ^{2} + y ^{2}}), y (1) = 0$ is :2023 · MCQ
Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . Then $(y (2))^{3} - 12 y (2)$ is equal to :2023 · MCQ
If x = x(y) is the solution of the differential equation $y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}, x (1) = 0$ ; then x(e) is equal to :2022 · MCQ
The slope of the tangent to a curve $C : y = y (x)$ at any point $\frac{2 e ^{2 x} - 6 e ^{- x} + 9}{2 + 9 e ^{- 2 x}}$ on it is $\frac{2 e ^{2 x} - 6 e ^{- x} + 9}{2 + 9 e ^{- 2 x}}$ . If $C$ passes through the points $$\left(0, \frac{1}{2}+\frac{\pi}{2…2022 · MCQ
The general solution of the differential equation $(x - y^{2}) d x + y (5 x + y^{2}) d y = 0$ is :2022 · MCQ
Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{4 y ^{3} + 2 y x ^{2}}{3 x y ^{2} + x ^{3}}, y (1) = 1$ . If for some $n \in N, y (2) \in [n - 1, n)$ , then $n$ is equal to .2022 · Numerical