Advertisements
Advertisements
प्रश्न
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
Advertisements
उत्तर
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
\[ \Rightarrow \left( 1 + y^2 \right) \tan^{- 1} xdx = - 2y\left( 1 + x^2 \right)dy\]
\[ \Rightarrow \frac{\tan^{- 1} x}{\left( 1 + x^2 \right)}dx = - \frac{2y}{\left( 1 + y^2 \right)}dy\]
\[ \Rightarrow \int\frac{\tan^{- 1} x}{\left( 1 + x^2 \right)}dx = - \int\frac{2y}{\left( 1 + y^2 \right)}dy\]
\[ \Rightarrow \frac{\left( \tan^{- 1} x \right)^2}{2} = - \log\left| 1 + y^2 \right| + C\]
\[ \Rightarrow \frac{\left( \tan^{- 1} x \right)^2}{2} + \log\left| 1 + y^2 \right| = C\]
APPEARS IN
संबंधित प्रश्न
Verify that y = 4 sin 3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 9y = 0\]
Verify that y = \[\frac{a}{x} + b\] is a solution of the differential equation
\[\frac{d^2 y}{d x^2} + \frac{2}{x}\left( \frac{dy}{dx} \right) = 0\]
Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]
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
Differential equation \[\frac{d^2 y}{d x^2} - y = 0, y \left( 0 \right) = 2, y' \left( 0 \right) = 0\] Function y = ex + e−x
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
(y2 − 2xy) dx = (x2 − 2xy) dy
2xy dx + (x2 + 2y2) dy = 0
(x + 2y) dx − (2x − y) dy = 0
Solve the following initial value problem:-
\[\frac{dy}{dx} + 2y = e^{- 2x} \sin x, y\left( 0 \right) = 0\]
Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.
Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] are rectangular hyperbola.
The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
The equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) 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 \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
Choose the correct alternative:
Differential equation of the function c + 4yx = 0 is
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
Solve: ydx – xdy = x2ydx.
