The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is
Step 1: Given Differential Equation
We are given the following differential equation:
\[ \frac{dy}{dx} = \frac{x + a}{y - 2} \] This is a separable equation that we can manipulate to solve for \( y \) in terms of \( x \).
Step 2: Rearranging the Equation
By multiplying both sides by \( (2 - y) \) and integrating, we get:
\[ (2 - y) \, dy = (x + a) \, dx \]
Step 3: Integrating \p>Now, we integrate both sides:
\[ 2y - y^2 = \frac{x^2}{2} + ax + c \] where \( a + c = -\frac{1}{2} \) as \( y(1) = 0 \). Thus, we get the equation:
\[ x^2 + y^2 + 2ax - 4y - 1 - 2a = 0 \]
Step 4: Solving for \( a \) and \( c \)
We now solve for \( a \) and \( c \) using the equation:
\[ \pi r^2 = 4\pi \quad \Rightarrow \quad r^2 = 4 \] Thus, the equation simplifies to: \[ 4 = \sqrt{a^2 + 4 + 1 + 2a} \] Which leads to: \[ (a + 1)^2 = 0 \quad \Rightarrow \quad a = -1 \]
Step 5: Finding the Points P and Q
We know that \( P, Q = (0, 2 \pm \sqrt{3}) \). The equation of the normal line at these points is:
\[ y - 2 = -\sqrt{3}(x - 1) \]
Step 6: Calculating the Lengths R, S, and RS
The points \( R \) and \( S \) are given as:
\[ R = \left( 1 - \frac{2}{\sqrt{3}}, 0 \right), \quad S = \left( 1 + \frac{2}{\sqrt{3}}, 0 \right) \] The distance \( RS \) is calculated as: \[ RS = \frac{4}{\sqrt{3}} \quad \Rightarrow \quad RS = 4\sqrt{3} \]
Conclusion \p>The final answer is found by calculating the length \( RS \), which is \( 4\sqrt{3} \).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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.
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’
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”.
Differential equations can be divided into several types namely