English

√ 1 + X 2 D Y + √ 1 + Y 2 D X = 0

Advertisements
Advertisements

Question

\[\sqrt{1 + x^2} dy + \sqrt{1 + y^2} dx = 0\]
Advertisements

Solution

We have,
\[\sqrt{1 + x^2} dy + \sqrt{1 + y^2} dx = 0\]
\[\sqrt{1 + x^2} dy = - \sqrt{1 + y^2} dx\]
\[\frac{1}{\sqrt{1 + y^2}}dy = - \frac{1}{\sqrt{1 + x^2}}dx\]
Integrating both sides, we get
\[\int\frac{1}{\sqrt{1 + y^2}}dy = - \int\frac{1}{\sqrt{1 + x^2}}dx\]
\[ \Rightarrow \log \left| y + \sqrt{1 + y^2} \right| = - \log \left| x + \sqrt{1 + x^2} \right| + \log C\]
\[ \Rightarrow \log \left| y + \sqrt{1 + y^2} \right| + \log \left| x + \sqrt{1 + x^2} \right| = \log C\]
\[ \Rightarrow \log \left| \left( y + \sqrt{1 + y^2} \right)\left( x + \sqrt{1 + x^2} \right) \right| = \log C\]
\[ \Rightarrow \left( y + \sqrt{1 + y^2} \right)\left( x + \sqrt{1 + x^2} \right) = C\]
\[\text{ Hence, }\log \left( y + \sqrt{1 + y^2} \right)\left( x + \sqrt{1 + x^2} \right) =\text{ C is the required differential equation .} \]

shaalaa.com
  Is there an error in this question or solution?
Chapter 21: Differential Equations - Exercise 22.07 [Page 55]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.07 | Q 17 | Page 55

RELATED QUESTIONS

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

Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].


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


\[\frac{dy}{dx} = x^2 + x - \frac{1}{x}, x \neq 0\]

\[\frac{dy}{dx} = x^5 + x^2 - \frac{2}{x}, x \neq 0\]

\[\frac{dy}{dx} = \tan^{- 1} x\]


(sin x + cos x) dy + (cos x − sin x) dx = 0


(ey + 1) cos x dx + ey sin x dy = 0


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

tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y) 

 


\[\frac{dy}{dx} + \frac{\cos x \sin y}{\cos y} = 0\]

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

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

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

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


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

\[2xy\frac{dy}{dx} = x^2 + y^2\]

Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]


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 \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]


Define a differential equation.


If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.


The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting


What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?


Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].


Form the differential equation representing the family of curves y = a sin (x + b), where ab are arbitrary constant.


Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2). 


Choose the correct option from the given alternatives:

The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is


Solve the following differential equation.

`y^3 - dy/dx = x dy/dx`


For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1


Choose the correct alternative.

The differential equation of y = `k_1 + k_2/x` is


The solution of `dy/dx + x^2/y^2 = 0` is ______


Solve the following differential equation y log y = `(log  y - x) ("d"y)/("d"x)`


A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution


Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.


If `y = log_2 log_2(x)` then `(dy)/(dx)` =


In mathematics and science, what primary purpose do differential equations serve?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×