Let the solution curve y = y(x) of the differential equation (4 + x2)dy – 2x(x2 + 3y + 4)dx = 0 pass through the origin. Then y(2) is equal to _______.
The correct answer is 12
(4 + x2) dy – 2x(x2 + 3y + 4)dx = 0
\(⇒\frac{dy}{dx}=(\frac{6x}{x^2+4})y+2x\)
\(⇒\frac{dy}{dx}−(\frac{6x}{x^2+4})y=2x\)
I.F. \(= e^{−3 In(x^2+4)}=\frac{1}{(x^2+4)^3}\)
\(So, \frac{y}{(x^2+4)^3}=∫\frac{2x}{(x^2+4)^3}dx+c\)
\(⇒y=−\frac{1}{2}(x^2+4)+c(x^2+4)^3\)
When x = 0, y = 0 gives
\(c=\frac{1}{32}\)
So, for x = 2,y = 12
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:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).
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