मराठी

Find the Solution of the Differential Equation X √ 1 + Y 2 D X + Y √ 1 + X 2 D Y = 0

Advertisements
Advertisements

प्रश्न

Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]

बेरीज
Advertisements

उत्तर

\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
\[ \Rightarrow y\sqrt{1 + x^2}dy = - x\sqrt{1 + y^2}dx\]
\[ \Rightarrow \frac{y}{\sqrt{1 + y^2}}dy = \frac{- x}{\sqrt{1 + x^2}}dx\]
\[ \Rightarrow \int\frac{y}{\sqrt{1 + y^2}}dy = - \int\frac{x}{\sqrt{1 + x^2}}dx\]
\[\text{Let }1 + y^2 = t^2\text{ and }1 + x^2 = p^2 \]
\[ \Rightarrow 2ydy = 2tdt\text{ and }2xdx = 2pdp\]
\[ \Rightarrow ydy = tdt\text{ and }xdx = pdp\]
Substituting in above equation, we get
\[ \Rightarrow \int dt = - \int dp\]
\[ \Rightarrow t = - p + C\]
\[ \Rightarrow \sqrt{1 + x^2} + \sqrt{1 + y^2} = C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 21: Differential Equations - Very Short Answers [पृष्ठ १३९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Very Short Answers | Q 30 | पृष्ठ १३९

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

\[y\frac{d^2 x}{d y^2} = y^2 + 1\]

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


Verify that y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]


Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]


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
\[y = \left( \frac{dy}{dx} \right)^2\]
\[y = \frac{1}{4} \left( x \pm a \right)^2\]

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


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

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

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

\[\frac{dy}{dx} = e^{x + y} + e^y x^3\]

Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]


\[2x\frac{dy}{dx} = 5y, y\left( 1 \right) = 1\]

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

Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.


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

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

(x + y) (dx − dy) = dx + dy


3x2 dy = (3xy + y2) dx


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

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

Solve the following initial value problem:-

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


The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.


A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.


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 equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.


Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.


The differential equation \[x\frac{dy}{dx} - y = x^2\], has the general solution


In each of the following examples, verify that the given function is a solution of the corresponding differential equation.

Solution D.E.
y = ex  `dy/ dx= y`

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`

For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1


For  the following differential equation find the particular solution.

`dy/ dx = (4x + y + 1),

when  y = 1, x = 0


Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`


Solve the following differential equation.

`dy/dx + y = e ^-x`


State whether the following is True or False:

The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.


Solve

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


An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.


Solve: ydx – xdy = x2ydx.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×