मराठी

√ 1 + X 2 D Y + √ 1 + Y 2 D X = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 22: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Exercise 22.07 | Q 17 | पृष्ठ ५५

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

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

Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.


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 \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]

Function y = ex + 1


Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x


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

\[\frac{dy}{dx} = x^5 \tan^{- 1} \left( x^3 \right)\]

\[\frac{dy}{dx} = \frac{1 - \cos 2y}{1 + \cos 2y}\]

\[5\frac{dy}{dx} = e^x y^4\]

tan y dx + sec2 y tan x dy = 0


\[\frac{dy}{dx} = 1 - x + y - xy\]

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

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

 


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.


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

3x2 dy = (3xy + y2) dx


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

 

\[\left[ x\sqrt{x^2 + y^2} - y^2 \right] dx + xy\ dy = 0\]

Solve the following initial value problem:-

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


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


Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

 

At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.


Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.


The solution of the differential equation y1 y3 = y22 is


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). 


Determine the order and degree of the following differential equations.

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

Determine the order and degree of the following differential equations.

Solution D.E.
ax2 + by2 = 5 `xy(d^2y)/dx^2+ x(dy/dx)^2 = y dy/dx`

Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0


Solve the following differential equation.

`x^2 dy/dx = x^2 +xy - y^2`


Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0


Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`


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


Solve: ydx – xdy = x2ydx.


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


Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×