Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[\frac{dy}{dx} + \frac{\cos x \sin y}{\cos y} = 0\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{\cos x \sin y}{\cos y}\]
\[ \Rightarrow \frac{\cos y}{\sin y}dy = - \cos x dx\]
\[ \Rightarrow \cot y dy = - \cos x dx\]
Integrating both sides, we get
\[\int\cot y dy = - \int\cos x dx\]
\[ \Rightarrow \log \left| \sin y \right| = - \sin x + C\]
\[\text{ Hence,} \log \left| \sin y \right| = - \sin x +\text{ C is the required solution .} \]
APPEARS IN
संबंधित प्रश्न
Prove that:
`int_0^(2a)f(x)dx = int_0^af(x)dx + int_0^af(2a - x)dx`
If 1, `omega` and `omega^2` are the cube roots of unity, prove `(a + b omega + c omega^2)/(c + s omega + b omega^2) = omega^2`
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} + y = y^2\]
|
\[y = \frac{a}{x + a}\]
|
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
(sin x + cos x) dy + (cos x − sin x) dx = 0
(1 + x2) dy = xy dx
xy (y + 1) dy = (x2 + 1) 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.
In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.
Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
Solve the differential equation:
`"x"("dy")/("dx")+"y"=3"x"^2-2`
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
The differential equation `y dy/dx + x = 0` represents family of ______.
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Solve the following differential equation.
`(dθ)/dt = − k (θ − θ_0)`
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
The solution of `dy/dx + x^2/y^2 = 0` is ______
Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
