Advertisements
Advertisements
प्रश्न
Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]
Advertisements
उत्तर
We have,
\[x\frac{dy}{dx} + \cot y = 0\]
\[ \Rightarrow x\frac{dy}{dx} = - \cot y\]
\[ \Rightarrow \tan y\ dy = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\tan y\ dy = - \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| \sec y \right| = - \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left( \left| x \right| \left| \sec y \right| \right) = \log C\]
\[ \Rightarrow x \sec y = C . . . . . (1) \]
\[\text{ Given:- } x = \sqrt{2}, y = \frac{\pi}{4} . \]
Substituting the values of x and y in (1), we get
\[\sqrt{2} sec \frac{\pi}{4} = C\]
\[ \Rightarrow C = 2\]
Substituting the value of C in (1), we get
\[x \sec y = 2\]
\[ \Rightarrow x = 2 \cos y\]
\[\text{Hence, }x = 2 \cos y\text{ is the required solution . }\]
APPEARS IN
संबंधित प्रश्न
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x^3 \frac{d^2 y}{d x^2} = 1\]
|
\[y = ax + b + \frac{1}{2x}\]
|
Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex
Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 1\] Function y = sin x + cos x
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 \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
2xy dx + (x2 + 2y2) dy = 0
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:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]
The slope of the tangent at a point P (x, y) on a curve is \[\frac{- x}{y}\]. If the curve passes through the point (3, −4), find the equation of the curve.
Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.
A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.
Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]
The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
y = ex + 1 y'' − y' = 0
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
`y=sqrt(a^2-x^2)` `x+y(dy/dx)=0`
The price of six different commodities for years 2009 and year 2011 are as follows:
| Commodities | A | B | C | D | E | F |
|
Price in 2009 (₹) |
35 | 80 | 25 | 30 | 80 | x |
| Price in 2011 (₹) | 50 | y | 45 | 70 | 120 | 105 |
The Index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x andy if the total price in 2009 is ₹ 360.
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
Solve the following differential equation.
x2y dx − (x3 + y3) dy = 0
Choose the correct alternative.
The differential equation of y = `k_1 + k_2/x` is
Solve the differential equation:
`e^(dy/dx) = x`
Solve:
(x + y) dy = a2 dx
x2y dx – (x3 + y3) dy = 0
Select and write the correct alternative from the given option for the question
Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in
Solve the differential equation xdx + 2ydy = 0
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
Solve the differential equation
`y (dy)/(dx) + x` = 0
