Advertisements
Advertisements
Question
Solution
We have,
\[\frac{dy}{dx} = \left( \cos^2 x - \sin^2 x \right) \cos^2 y\]
\[ \Rightarrow \frac{dy}{dx} = \cos 2x \cos^2 y\]
\[ \Rightarrow \frac{1}{\cos^2 y}dy = \cos 2x dx\]
\[ \Rightarrow \sec^2 y dy = \cos 2x dx\]
Integrating both sides, we get
\[\int \sec^2 y dy = \int\cos 2x dx\]
\[ \Rightarrow \tan y = \frac{\sin 2x}{2} + C\]
\[\text{ Hence, }\tan y = \frac{\sin 2x}{2} +\text{ C is the required solution }.\]
APPEARS IN
RELATED QUESTIONS
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 \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]
Function y = log x
xy (y + 1) dy = (x2 + 1) dx
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
x2 dy + y (x + y) dx = 0
(x + 2y) dx − (2x − y) dy = 0
In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?
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).
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.
If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.
The solution of the differential equation y1 y3 = y22 is
The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is
The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]
y2 dx + (x2 − xy + y2) dy = 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
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
Determine the order and degree of the following differential equations.
| Solution | D.E. |
| y = 1 − logx | `x^2(d^2y)/dx^2 = 1` |
Form the differential equation from the relation x2 + 4y2 = 4b2
For the following differential equation find the particular solution.
`(x + 1) dy/dx − 1 = 2e^(−y)`,
when y = 0, x = 1
The solution of `dy/ dx` = 1 is ______
Solve:
(x + y) dy = a2 dx
y2 dx + (xy + x2)dy = 0
`xy dy/dx = x^2 + 2y^2`
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
Solve the differential equation xdx + 2ydy = 0
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.
