Let the solution curve y = y(x) of the differential equation
\([ \frac{x}{\sqrt{x² -y²}} + e^\frac{y}{x} ] x \frac{dy}{dx} = x + [ \frac{x}{\sqrt{x² -y²}} + e^\frac{y}{x} ]y\)
pass through the points (1, 0) and (2α, α), α> 0. Then α is equal to
Let the solution curve y = y(x) of the differential equation
\([ \frac{x}{\sqrt{x² -y²}} + e^\frac{y}{x} ] x \frac{dy}{dx} = x + [ \frac{x}{\sqrt{x² -y²}} + e^\frac{y}{x} ]y\)
pass through the points (1, 0) and (2α, α), α> 0. Then α is equal to
\(\frac{1}{2} exp ( \frac{π}{6} + \sqrt{e} - 1 )\)
\(\frac{1}{2} exp ( \frac{π}{3} + e - 1 )\)
\(exp ( \frac{π}{6} + \sqrt{e} + 1 )\)
\(2 exp ( \frac{π}{3} + \sqrt{e} - 1 )\)
The Correct Option is A
Solution and Explanation
The correct answer is (A) : \(\frac{1}{2} exp ( \frac{π}{6} + \sqrt{e} - 1 )\)
\(\left( \frac{1}{ \sqrt{1 - \frac{y²}{x²}}} + e^\frac{y}{x} \right) \frac{dy}{dx} = 1 + \left( \frac{1}{\sqrt{1 - \frac{y²}{x²}}} + e^\frac{y}{x} \right) \frac{y}{x}\)
Putting y = tx
\(\left( \frac{1}{\sqrt{1 - t²}} + e^t \right) \left( t + x \frac{dt}{dx} \right) = 1 + \left( \frac{1}{\sqrt{1 - t²}} + e^t \right)t\)
\(⇒ x ( \frac{1}{\sqrt{1 - t²}} + e^t ) \frac{dt}{dx} = 1\)
⇒ sin–1t + et = lnx + C
\(⇒sin^{-1} ( \frac{y}{x} ) + e^\frac{y}{x} = In x + C\)
at x = 1, y = 0
So, 0 + e0 = 0 + C⇒C = 1
at (2α, α)
\(sin^{-1} ( \frac{y}{x} ) + e^\frac{y}{x} = In x + 1\)
\(⇒ \frac{π}{6} + e^\frac{1}{2} - 1 = In (2α)\)
\(⇒ α = \frac{1}{2} e^{( \frac{π}{6} + e^\frac{1}{2} - 1 )}\)
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- Let \( A \) be a \( 3 \times 3 \) matrix such that \( X^T AX = 0 \) for all nonzero \( 3 \times 1 \) matrices \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \). \[ A = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}, \quad A = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix} \] If \( \det(\text{adj}(2(A + I))) = 2^{\alpha} 3^{\beta} 5^{\gamma} \), \( \alpha, \beta, \gamma \in \mathbb{N} \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is:
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Concepts Used:
Differential Equations
A differential equation is an equation that contains one or more functions with its derivatives. The derivatives of the function define the rate of change of a function at a point. It is mainly used in fields such as physics, engineering, biology and so on.
Orders of a Differential Equation
First Order Differential Equation
The first-order differential equation has a degree equal to 1. All the linear equations in the form of derivatives are in the first order. It has only the first derivative such as dy/dx, where x and y are the two variables and is represented as: dy/dx = f(x, y) = y’
Second-Order Differential Equation
The equation which includes second-order derivative is the second-order differential equation. It is represented as; d/dx(dy/dx) = d2y/dx2 = f”(x) = y”.
Types of Differential Equations
Differential equations can be divided into several types namely
- Ordinary Differential Equations
- Partial Differential Equations
- Linear Differential Equations
- Nonlinear differential equations
- Homogeneous Differential Equations
- Nonhomogeneous Differential Equations