हिंदी

X Cos2 Y Dx = Y Cos2 X Dy - Mathematics

Advertisements
Advertisements

प्रश्न

x cos2 y  dx = y cos2 x dy

योग
Advertisements

उत्तर

We have,
\[x \cos {}^2 y dx = y \cos {}^2 x dy\]
\[ \Rightarrow \frac{x}{\cos^2 x}dx = \frac{y}{\cos^2 y}dy\]
\[ \Rightarrow x \sec^2 x dx = y \sec^2 y dy\]
Integrating both sides, we get 

\[ \Rightarrow x\int \sec^2 x dx - \int\left\{ \frac{d}{dx}\left( x \right)\int \sec^2 x dx \right\}dx = y\int \sec^2 y dy - \int\left\{ \frac{d}{dy}\left( y \right)\int \sec^2 y dy \right\}dy\]
\[ \Rightarrow x \tan x - \int\tan x dx = y \tan y - \int\tan y dy\]
\[ \Rightarrow x \tan x - \log \left| \sec x \right| = y \tan y - \log \left| \sec y \right| + C\]
\[ \Rightarrow x \tan x - y \tan y = \log \left| \sec x \right| - \log \left| \sec y \right| + C\]
\[\text{ Hence, }x \tan x - y \tan y = \log \left| \sec x \right| - \log \left| \sec y \right| +\text{C is the required solution.} \]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 22: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 22 Differential Equations
Exercise 22.07 | Q 11 | पृष्ठ ५५

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Solve the equation for x: `sin^(-1)  5/x + sin^(-1)  12/x = π/2, x ≠ 0`


\[\frac{d^3 x}{d t^3} + \frac{d^2 x}{d t^2} + \left( \frac{dx}{dt} \right)^2 = e^t\]

\[\frac{d^2 y}{d x^2} + 4y = 0\]

\[\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \left( c\frac{d^2 y}{d x^2} \right)^{1/3}\]

\[x^2 \left( \frac{d^2 y}{d x^2} \right)^3 + y \left( \frac{dy}{dx} \right)^4 + y^4 = 0\]

Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.


Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.


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 + ex


\[\left( x^3 + x^2 + x + 1 \right)\frac{dy}{dx} = 2 x^2 + x\]

(1 + x2) dy = xy dx


\[\frac{dy}{dx} = \left( e^x + 1 \right) y\]

\[\left( x - 1 \right)\frac{dy}{dx} = 2 x^3 y\]

x cos y dy = (xex log x + ex) dx


(ey + 1) cos x dx + ey sin x dy = 0


\[\frac{dy}{dx} = \frac{e^x \left( \sin^2 x + \sin 2x \right)}{y\left( 2 \log y + 1 \right)}\]

\[\frac{dy}{dx} = 2xy, y\left( 0 \right) = 1\]

Find the particular solution of edy/dx = x + 1, given that y = 3, when x = 0.


The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after `t` seconds.


\[\frac{dy}{dx} = \frac{\left( x - y \right) + 3}{2\left( x - y \right) + 5}\]

\[\left( x + y \right)^2 \frac{dy}{dx} = 1\]

Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]


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.


The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).


Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.


The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is


The solution of the differential equation y1 y3 = y22 is


The differential equation satisfied by ax2 + by2 = 1 is


Which of the following differential equations has y = C1 ex + C2 ex as the general solution?


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 particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0


Determine the order and degree of the following differential equations.

Solution D.E.
y = 1 − logx `x^2(d^2y)/dx^2 = 1`

Solve the following differential equation.

(x2 − y2 ) dx + 2xy dy = 0


The integrating factor of the differential equation `dy/dx - y = x` is e−x.


Solve the differential equation:

`e^(dy/dx) = x`


 `dy/dx = log x`


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`


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 = – 2c

This is general solution.

When x = `pi/6`, y = 0, we have

`"e"^0 + 2sin  pi/6` = c

∴ c = `square`

∴ particular solution is `square`


Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×