Homogeneous Differential Equations

Interactive Feed

JEE Main 2008

The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

y = ln x + x

(B)

y = x ln x + x^{2}

(C)

y = x e^{(x - 1)}

(D)

y = x ln x + x

JEE Main 2005

LEVELJEE Main

If $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , then the solution of the equation is

(A)

y lo g (\frac{x}{y}) = c x

(B)

x lo g (\frac{y}{x}) = cy

(C)

lo g (\frac{y}{x}) = c x

(D)

lo g (\frac{x}{y}) = cy

JEE Advanced 2013

LEVELJEE Main

A curve passes through the point $(1, π /6)$ . Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + sec (\frac{y}{x}), x > 0$ . Then the equation of the curve is

(A)

sin (\frac{y}{x}) = ln x + \frac{1}{2}

(B)

csc (\frac{y}{x}) = ln x + 2

(C)

sec (\frac{2 y}{x}) = ln x + 2

(D)

cos (\frac{2 y}{x}) = ln x + \frac{1}{2}

JEE Advanced 2005

LEVELJEE Main

The solution of primitive integral equation $(x^{2} + y^{2}) d y = x y d x$ is $y = y (x)$ . If $y (1) = 1$ and $y (x_{0}) = e$ , then $x_{0}$ is equal to

(A)

2 (e^{2} - 1)

(B)

2 (e^{2} + 1)

(C)

3 e

(D)

\frac{e ^{2} + 1}{2}

JEE Advanced 1999

LEVELJEE Advanced

A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve.

JEE Main 2018 (15 April Evening)

LEVELJEE Main

The curve satisfying the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ and passing through the point (1, 1) is :-

(A)

A hyperbola

(B)

A circle of radius two

(C)

A circle of radius one

(D)

An ellipse

JEE Main 2019 (10 January)

LEVELJEE Main

The curve amongst the family of curves, represented by the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ which passes through $(1, 1)$ is :

(A)

A circle with centre on the y-axis

(B)

A circle with centre on the x-axis

(C)

An ellipse with major axis along the y-axis

(D)

A hyperbola with transverse axis along the x-axis

JEE Main 2020 - 9 Jan (Evening)

LEVELJEE Main

If $\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}$ ; $y (1) = 1$ : then a value of $x$ satisfying $y (x) = e$ is:

(A)

3 e

(B)

\frac{e}{2}

(C)

2 e

(D)

\frac{3}{2} e

JEE Main 2020 - 2 Sep (Evening)

LEVELJEE Main

If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :

(A)

\frac{1}{1 + l o g _{e} 2}

(B)

\frac{- 1}{1 + l o g _{e} 2}

(C)

1 + lo g_{e} 2

(D)

\frac{1}{1 - l o g _{e} 2}

JEE Main 2021 (20 July Shift 1)

LEVELJEE Advanced

Let $y = y (x)$ be the solution of the differential equation $x tan (\frac{y}{x}) d y = (y tan (\frac{y}{x}) - x) d x$ , $- 1 \leq x \leq 1, y (\frac{1}{2}) = \frac{π}{6}$ . Then the area of the region bounded by the curves $x = 0, x = \frac{1}{2}$ and $y = y (x)$ in the upper half plane is:

(A)

\frac{1}{8} (π - 1)

(B)

\frac{1}{12} (π - 3)

(C)

\frac{1}{4} (π - 2)

(D)

\frac{1}{6} (π - 1)

JEE Main 2021 (31 August Shift 2)

LEVELJEE Main

If $y \frac{d y}{d x} = x [\frac{y ^{2}}{x ^{2}} + \frac{ϕ ( \frac{y ^{2}}{x ^{2}} )}{ϕ ^{'} ( \frac{y ^{2}}{x ^{2}} )}], x > 0, ϕ > 0$ , and $y (1) = - 1$ , then $ϕ (\frac{y ^{2}}{4})$ is equal to :

(A)

4 ϕ (2)

(B)

4 ϕ (1)

(C)

2 ϕ (1)

(D)

ϕ (1)

JEE Main 2021 (16 March Shift 2)

LEVELJEE Advanced

Let $C_{1}$ be the curve obtained by the solution of differential equation $2 x y \frac{d y}{d x} = y^{2} - x^{2}, x > 0$ & Let the curve $C_{2}$ be the solution of $\frac{2 x y}{x ^{2} - y ^{2}} = \frac{d y}{d x}$ . If both the curves pass through $(1, 1)$ , then the area enclosed by the curves $C_{1}$ and $C_{2}$ is equal to :

(A)

π - 1

(B)

\frac{π}{2} - 1

(C)

π + 1

(D)

\frac{π}{4} + 1

JEE Main 2021 (18 March Shift 2)

LEVELJEE Advanced

Let $y = y (x)$ be the solution of the differential equation $x d y - y d x = x^{2} - y^{2} d x, x \geq 1$ , with $y (1) = 0$ . If the area bounded by the line $x = 1, x = e^{π}, y = 0$ and $y = y (x)$ is $α e^{2 π} + β$ , then the value of $10 (α + β)$ is equal to ____.

JEE Main 2022 (25 June Shift 2)

LEVELJEE Main

If $y = y (x)$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that $y (e) = \frac{e}{3}$ , then $y (1)$ is equal to

(A)

\frac{1}{3}

(B)

\frac{2}{3}

(C)

\frac{3}{2}

(D)

JEE Main 2022 (28 June Shift 1)

LEVELJEE Main

Let the solution curve $y = y (x)$ of the differential equation, $[\frac{x}{x ^{2} - y ^{2}} + e^{y / x}] x \frac{d y}{d x} = x + [\frac{x}{x ^{2} - y ^{2}} + e^{y / x}] y$ pass through the points $(1, 0)$ and $(2 α, α), α > 0$ . Then $α$ is equal to

(A)

\frac{1}{2} exp (\frac{π}{6} + e - 1)

(B)

\frac{1}{2} exp (\frac{π}{3} + e - 1)

(C)

exp (\frac{π}{6} + e + 1)

(D)

2 exp (\frac{π}{3} + e - 1)

JEE Main 2022 (29 June Shift 1)

LEVELJEE Main

Let the solution curve of the differential equation $x \frac{d y}{d x} - y = y^{2} + 16 x^{2}, y (1) = 3$ be $y = y (x)$ . Then $y (2)$ is equal to :

JEE Main 2022 (25 July Shift 2)

LEVELJEE Main

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

JEE Main 2023 (24 January Shift 2)

LEVELJEE Main

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ is equal to

(A)

3 e^{2}

(B)

e^{2}

(C)

2 e^{2}

(D)

\frac{3 e ^{2}}{2}

JEE Main 2023 (30 January Shift 2)

LEVELJEE Main

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

(A)

lo g_{e} ∣ x + y ∣ - \frac{x y}{( x + y ) ^{2}} = 0

(B)

lo g_{e} ∣ x + y ∣ + \frac{x y}{( x + y ) ^{2}} = 0

(C)

lo g_{e} ∣ x + y ∣ + \frac{2 x y}{( x + y ) ^{2}} = 0

(D)

lo g_{e} ∣ x + y ∣ - \frac{2 x y}{( x + y ) ^{2}} = 0

JEE Main 2023 (31 January Shift 2)

LEVELJEE Main

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:

(A)

322

(B)

(C)

162

(D)

JEE Main 2024 (09 April Shift 1)

LEVELJEE Main

The solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0, y (1) = 0$ , is :

(A)

∣ x^{2} - 2 y^{2} ∣^{6} = x

(B)

∣ x^{2} - 4 y^{2} ∣^{6} = x

(C)

∣ x^{2} - 4 y^{2} ∣^{5} = x^{2}

(D)

∣ x^{2} - 2 y^{2} ∣^{5} = x^{2}

JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Advanced

If the solution curve, of the differential equation $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$ passing through the point $(2, 1)$ is $tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} lo g_{e} (α + (\frac{y - 1}{x - 1})^{2}) = lo g_{e} ∣ x - 1∣$ , then $5 β + α$ is equal to

JEE Main 2024 (29 Jan Shift 2)

LEVELJEE Main

If $sin (\frac{y}{x}) = lo g_{e} ∣ x ∣ + \frac{α}{2}$ is the solution of the differential equation $x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to

JEE Main 2024 (31 Jan Shift 1)

LEVELJEE Main

The solution curve of the differential equation $y \frac{d x}{d y} = x (lo g_{e} x - lo g_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

(A)

∣ lo g_{e} \frac{x}{y} ∣ = x

(B)

∣ lo g_{e} \frac{y}{x} ∣ = y^{2}

(C)

∣ lo g_{e} \frac{x}{y} ∣ = y

(D)

2∣ lo g_{e} \frac{x}{y} ∣ = y + 1

JEE Main 2026 (22 January Shift 1)

LEVELJEE Main

Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation dxdy​=xx+y​ satisfying the condition y(1)=1 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If xdxdy​=y(logy−logx+1), then the solution of the equation is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A curve passes through the point (1,π/6). Let the slope of the curve at each point (x,y) be xy​+sec(xy​),x>0. Then the equation of the curve is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of primitive integral equation (x2+y2)dy=xydx is y=y(x). If y(1)=1 and y(x0​)=e, then x0​ is equal to

A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The curve satisfying the differential equation, (x2−y2)dx+2xydy=0 and passing through the point (1, 1) is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The curve amongst the family of curves, represented by the differential equation, (x2−y2)dx+2xydy=0 which passes through (1,1) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=x2+y2xy​; y(1)=1: then a value of x satisfying y(x)=e is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a curve y=f(x), passing through the point (1,2), is the solution of the differential equation, 2x2dy=(2xy+y2)dx, then f(21​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ydxdy​=x[x2y2​+ϕ′(x2y2​)ϕ(x2y2​)​],x>0,ϕ>0, and y(1)=−1, then ϕ(4y2​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation 2x2dxdy​−2xy+3y2=0 such that y(e)=3e​, then y(1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the solution curve of the differential equation xdxdy​−y=y2+16x2​,y(1)=3 be y=y(x). Then y(2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​=3xy2+x34y3+2yx2​,y(1)=1. If for some n∈N,y(2)∈[n−1,n), then n is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation (x2−3y2)dx+3xydy=0,y(1)=1. Then 6y2(e) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation dxdy​=−(3x2+y2x2+3y2​),y(1)=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation (3y2−5x2)ydx+2x(x2−y2)dy=0 such that y(1)=1. then ∣(y(2))3−12y(2)∣ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation (x2+y2)dx−5xydy=0,y(1)=0, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If sin(xy​)=loge​∣x∣+2α​ is the solution of the differential equation xcos(xy​)dxdy​=ycos(xy​)+x and y(1)=3π​, then α2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution curve of the differential equation ydydx​=x(loge​x−loge​y+1),x>0,y>0 passing through the point (e,1) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the solution curve of the differential equation xdy−ydx=x2+y2​dx,x>0,y(1)=0, be y=y(x). Then y(3) is equal to

The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

If $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , then the solution of the equation is

A curve passes through the point $(1, π /6)$ . Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + sec (\frac{y}{x}), x > 0$ . Then the equation of the curve is

The solution of primitive integral equation $(x^{2} + y^{2}) d y = x y d x$ is $y = y (x)$ . If $y (1) = 1$ and $y (x_{0}) = e$ , then $x_{0}$ is equal to

The curve satisfying the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ and passing through the point (1, 1) is :-

The curve amongst the family of curves, represented by the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ which passes through $(1, 1)$ is :

If $\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}$ ; $y (1) = 1$ : then a value of $x$ satisfying $y (x) = e$ is:

If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :

If $y \frac{d y}{d x} = x [\frac{y ^{2}}{x ^{2}} + \frac{ϕ ( \frac{y ^{2}}{x ^{2}} )}{ϕ ^{'} ( \frac{y ^{2}}{x ^{2}} )}], x > 0, ϕ > 0$ , and $y (1) = - 1$ , then $ϕ (\frac{y ^{2}}{4})$ is equal to :

If $y = y (x)$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that $y (e) = \frac{e}{3}$ , then $y (1)$ is equal to

Let the solution curve of the differential equation $x \frac{d y}{d x} - y = y^{2} + 16 x^{2}, y (1) = 3$ be $y = y (x)$ . Then $y (2)$ is equal to :

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

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ is equal to

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

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:

The solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0, y (1) = 0$ , is :

If $sin (\frac{y}{x}) = lo g_{e} ∣ x ∣ + \frac{α}{2}$ is the solution of the differential equation $x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to

The solution curve of the differential equation $y \frac{d x}{d y} = x (lo g_{e} x - lo g_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to