Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[\frac{dy}{dx} = \sin^2 y\]
\[ \Rightarrow \frac{dx}{dy} = \frac{1}{\sin^2 y}\]
\[ \Rightarrow dx = {cosec}^2 y dy\]
Integrating both sides, we get
\[\int dx = \int {cosec}^2 y dy\]
\[ \Rightarrow x = - \cot y + C\]
\[ \Rightarrow x + \cot y = C\]
\[\text{ Hence, }x + \cot y = \text{ C is the required solution }.\]
APPEARS IN
संबंधित प्रश्न
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.
Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[y = \left( \frac{dy}{dx} \right)^2\]
|
\[y = \frac{1}{4} \left( x \pm a \right)^2\]
|
Differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]
Function y = ex + 1
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
(ey + 1) cos x dx + ey sin x dy = 0
(y + xy) dx + (x − xy2) dy = 0
Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\]
(y2 − 2xy) dx = (x2 − 2xy) dy
Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]
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:-
\[\frac{dy}{dx} + 2y \tan x = \sin x; y = 0\text{ when }x = \frac{\pi}{3}\]
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.
The differential equation satisfied by ax2 + by2 = 1 is
The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]
y2 dx + (x2 − xy + y2) dy = 0
Solve the following differential equation.
xdx + 2y dx = 0
Solve the following differential equation.
`dy/dx + 2xy = x`
The solution of `dy/ dx` = 1 is ______.
y dx – x dy + log x dx = 0
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve the differential equation xdx + 2ydy = 0
Solve: `("d"y)/("d"x) + 2/xy` = x2
Choose the correct alternative:
Differential equation of the function c + 4yx = 0 is
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
