The solution curve, of the differential equation \( 2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx} \), passing through the point \( (0, 1) \), is a conic, whose vertex lies on the line:
- \( 2x + 3y = 9 \)
- \( 2x + 3y = -9 \)
- \( 2x + 3y = -6 \)
- \( 2x + 3y = 6 \)
The Correct Option is A
Approach Solution - 1
To solve the given differential equation and find the line on which the vertex of the resulting conic lies, let's start by simplifying and solving the given differential equation:
The differential equation provided is: \(2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx}\)
Rearrange the terms to separate the variables:
\(2y \frac{dy}{dx} - 5 \frac{dy}{dx} = -3\)
Factor out \(\frac{dy}{dx}\):
\(\left( 2y - 5 \right) \frac{dy}{dx} = -3\)
Solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{-3}{2y - 5}\)
This is a separable differential equation. To solve this, separate the variables:
\((2y - 5) \, dy = -3 \, dx\)
Integrate both sides:
\(\int (2y - 5) \, dy = -\int 3 \, dx\)
This gives:
\(y^2 - 5y = -3x + C\)
Rearrange to get the equation of the conic:
\(y^2 - 5y + 3x = C\)
Use the initial condition that the curve passes through the point \((0, 1)\) to find the constant:
Substitute \((x, y) = (0, 1)\) in the equation:
\(1^2 - 5 \times 1 + 3 \times 0 = C\)
Calculate:
\(1 - 5 = C \Rightarrow C = -4\)
Thus, the equation of the conic becomes:
\(y^2 - 5y + 3x = -4\)
This is the equation of a conic. To find the vertex, consider completing the square for the equation in terms of \(y\):
\(y^2 - 5y = -(3x + 4)\)
Complete the square for the \(y\) terms:
\(y^2 - 5y + \left(\frac{5}{2}\right)^2 = \left(\frac{5}{2}\right)^2 - (3x + 4)\)
Which simplifies to:
\(\left(y - \frac{5}{2}\right)^2 = \frac{25}{4} - 3x - 4\)
Further simplify:
\(\left(y - \frac{5}{2}\right)^2 = \frac{9}{4} - 3x\)
The vertex of this conic section is at \((x, y) = (h, \frac{5}{2})\). For the equation to be zero at the vertex:
\(\frac{9}{4} = 3h\)
Solve for \(h\):
\(h = \frac{3}{4}\)
Thus, the vertex lies on the line \(y = \frac{5}{2}\).
Plug known values into the equation of the line to cross-check:
- \(2x + 3y = 9\) is tested for \((x, y) = \left(\frac{3}{4}, \frac{5}{2}\right)
Substitute \(x = \frac{3}{4}, y = \frac{5}{2}\):
\(2 \times \frac{3}{4} + 3 \times \frac{5}{2} = 9\)
Calculate:
\(\frac{3}{2} + \frac{15}{2} = 9\)
\(9 = 9\)
Thus, the correct line on which the vertex lies is \(2x + 3y = 9\).
Approach Solution -2
Solution:
\((2y - 5)\frac{dy}{dx} = -3\)
\((2y - 5)dy = -3dx\)
\(2\cdot\frac{y^2}{2} - 5y = -3x + \lambda\)
Given: Curve passes through \((0, 1)\)
\(\implies \lambda = -4\)
Curve Equation:
\(\left(\frac{y - 5}{2}\right)^2 = -3\left(x - \frac{3}{4}\right)\)
Vertex of the Parabola:
\(\left(\frac{3}{4}, \frac{5}{2}\right)\)
Final Equation: \(2x + 3y = 9\)
Top Questions on Differential equations
- Let \( y = y(x) \) be the solution of the differential equation \( \frac{dy}{dx} + 3(\tan^2 x) y + 3y = \sec^2 x \), with \( y(0) = \frac{1}{3} + e^3 \). Then \( y\left(\frac{\pi}{4}\right) \) is equal to
- JEE Main - 2025
- Mathematics
- Differential equations
- Let \( y = y(x) \) be the solution of the differential equation \[ \cos(x \log(\cos x))^2 \, dy + (\sin x - 3 \sin x \log(\cos x)) \, dx = 0, \quad x \in \left( 0, \frac{\pi}{2} \right) \] If \( y\left( \frac{\pi}{4} \right) = -1 \), then \( y\left( \frac{\pi}{6} \right) \) is equal to:
- JEE Main - 2025
- Mathematics
- Differential equations
- Let g be a differentiable function such that $ \int_0^x g(t) dt = x - \int_0^x tg(t) dt $, $ x \ge 0 $ and let $ y = y(x) $ satisfy the differential equation $ \frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x) $, $ x \in \left[ 0, \frac{\pi}{2} \right) $. If $ y(0) = 0 $, then $ y\left( \frac{\pi}{3} \right) $ is equal to
- JEE Main - 2025
- Mathematics
- Differential equations
- If a curve $ y = y(x) $ passes through the point $ \left(1, \frac{\pi}{2}\right) $ and satisfies the differential equation $$ (7x^4 \cot y - e^x \csc y) \frac{dx}{dy} = x^5, \quad x \geq 1, \text{ then at } x = 2, \text{ the value of } \cos y \text{ is:} $$
- JEE Main - 2025
- Mathematics
- Differential equations
- Let \( f(x) \) be a real differentiable function such that \( f(0) = 1 \) and \( f(x + y) = f(x)f'(y) + f'(x)f(y) \) for all \( x, y \in \mathbb{R} \). Then \( \sum_{n=1}^{100} \log_e f(n) \) is equal to :
- JEE Main - 2025
- Mathematics
- Differential equations
Questions Asked in JEE Main exam
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____.
- JEE Main - 2025
- Magnetic Field
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- JEE Main - 2025
- Solutions
- A physical quantity \( Q \) is related to four observables \( a \), \( b \), \( c \), and \( d \) as follows: \[ Q = \frac{a b^4}{c d^2} \] Where:
- \( a = (60 \pm 3) \, \text{Pa} \),
- \( b = (20 \pm 0.1) \, \text{m} \),
- \( c = (40 \pm 0.2) \, \text{N·s/m}^2 \),
- \( d = (50 \pm 0.1) \, \text{m} \).
Then the percentage error in \( Q \) is:- JEE Main - 2025
- Error analysis
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: - Assume a living cell with 0.9%(\(w/w\)) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data up to first decimal place only) The cell will:
- JEE Main - 2025
- osmosis