Advertisements
Advertisements
प्रश्न
tan y dx + sec2 y tan x dy = 0
Advertisements
उत्तर
We have,
\[\tan y dx + \sec^2 y \tan x dy = 0\]
\[ \Rightarrow \sec^2 y \tan x dy = - \tan y dx\]
\[ \Rightarrow \frac{\sec^2 y}{\tan y} dy = - \frac{1}{\tan x}dx\]
\[ \Rightarrow \frac{1}{\cos^2 y} \times \frac{\cos y}{\sin y}dy = - \cot x dx\]
\[ \Rightarrow \frac{1}{\sin y \cos y}dy = - \cot x dx\]
\[ \Rightarrow \frac{2}{\sin 2y}dy = - \cot x dx\]
\[ \Rightarrow 2 \text{ cosec } 2y dy = - \cot x dx\]
Integrating both sides, we get
\[2\int\text{ cosec }2y dy = - \int\cot x dx\]
\[ \Rightarrow \log \tan x = - \log \sin x = \log C\]
\[ \Rightarrow \log \tan x + \log \sin x = \log C\]
\[ \Rightarrow \log \left( \tan x \times \sin x \right) = \log C\]
\[ \Rightarrow \tan x \times \sin x = C\]
APPEARS IN
संबंधित प्रश्न
Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]
Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\] satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x + y\frac{dy}{dx} = 0\]
|
\[y = \pm \sqrt{a^2 - x^2}\]
|
Differential equation \[\frac{d^2 y}{d x^2} - y = 0, y \left( 0 \right) = 2, y' \left( 0 \right) = 0\] Function y = ex + e−x
Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex
C' (x) = 2 + 0.15 x ; C(0) = 100
(ey + 1) cos x dx + ey sin x dy = 0
(y + xy) dx + (x − xy2) dy = 0
x2 dy + y (x + y) dx = 0
Solve the following initial value problem:-
\[y' + y = e^x , y\left( 0 \right) = \frac{1}{2}\]
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x, y\left( 0 \right) = 1\]
Solve the following initial value problem:-
\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]
The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
Find the equation of the curve which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\] at any point (x, y) on it.
Define a differential equation.
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Find the differential equation whose general solution is
x3 + y3 = 35ax.
Solve the following differential equation.
`(dθ)/dt = − k (θ − θ_0)`
Solve the following differential equation.
`dy /dx +(x-2 y)/ (2x- y)= 0`
Choose the correct alternative.
Bacteria increases at the rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in
y2 dx + (xy + x2)dy = 0
`dy/dx = log x`
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
Find the particular solution of the following differential equation
`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.
Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x
∴ `1/"e"^(2y) "d"y` = cos x dx
Integrating, we get
`int square "d"y` = cos x dx
∴ `("e"^(-2y))/(-2)` = sin x + c1
∴ e–2y = – 2sin x – 2c1
∴ `square` = c, where c = – 2c1
This is general solution.
When x = `pi/6`, y = 0, we have
`"e"^0 + 2sin pi/6` = c
∴ c = `square`
∴ particular solution is `square`
lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is
