English

(X3 − 2y3) Dx + 3x2 Y Dy = 0

Advertisements
Advertisements

Question

(x3 − 2y3) dx + 3x2 y dy = 0

Sum
Advertisements

Solution

.....

shaalaa.com
  Is there an error in this question or solution?
Chapter 21: Differential Equations - Revision Exercise [Page 146]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Revision Exercise | Q 48 | Page 146

RELATED QUESTIONS

Solve the differential equation:  `x+ydy/dx=sec(x^2+y^2)` Also find the particular solution if x = y = 0.


If x = Φ(t) differentiable function of ‘ t ' then prove that `int f(x) dx=intf[phi(t)]phi'(t)dt`


Find the particular solution of differential equation:

`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`


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


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = ex + 1  :  y″ – y′ = 0


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = cos x + C : y′ + sin x = 0


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y – cos y = x :  (y sin y + cos y + x) y′ = y


Find the differential equation of the family of concentric circles `x^2 + y^2 = a^2`


If m and n are the order and degree of the differential equation \[\left( y_2 \right)^5 + \frac{4 \left( y_2 \right)^3}{y_3} + y_3 = x^2 - 1\], then


The solution of the differential equation \[\frac{dy}{dx} - ky = 0, y\left( 0 \right) = 1\] approaches to zero when x → ∞, if


The number of arbitrary constants in the particular solution of a differential equation of third order is


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


`y sec^2 x + (y + 7) tan x(dy)/(dx)=0`


`x cos x(dy)/(dx)+y(x sin x + cos x)=1`


`2 cos x(dy)/(dx)+4y sin x = sin 2x," given that "y = 0" when "x = pi/3.`


Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.


For the following differential equation, find the general solution:- \[\frac{dy}{dx} + y = 1\]


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


Solve the following differential equation:-

y dx + (x − y2) dy = 0


Find a particular solution of the following differential equation:- \[\left( 1 + x^2 \right)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}; y = 0,\text{ when }x = 1\]


Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x2 + 1) dx, x ≠ 0.


Solve the differential equation : `("x"^2 + 3"xy" + "y"^2)d"x" - "x"^2 d"y" = 0  "given that"  "y" = 0  "when"  "x" = 1`.


The solution of the differential equation `x "dt"/"dx" + 2y` = x2 is ______.


The general solution of the differential equation `"dy"/"dx" + y/x` = 1 is ______.


If y(x) is a solution of `((2 + sinx)/(1 + y))"dy"/"dx"` = – cosx and y (0) = 1, then find the value of `y(pi/2)`.


Solve: `y + "d"/("d"x) (xy) = x(sinx + logx)`


Integrating factor of `(x"d"y)/("d"x) - y = x^4 - 3x` is ______.


Solution of `("d"y)/("d"x) - y` = 1, y(0) = 1 is given by ______.


The solution of the differential equation cosx siny dx + sinx cosy dy = 0 is ______.


The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.


Solution of the differential equation `("d"y)/("d"x) + y/x` = sec x is ______.


The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is ______.


The solution of `("d"y)/("d"x) = (y/x)^(1/3)` is `y^(2/3) - x^(2/3)` = c.


The solution of the differential equation `("d"y)/("d"x) = (x + 2y)/x` is x + y = kx2.


Find the particular solution of the following differential equation, given that y = 0 when x = `pi/4`.

`(dy)/(dx) + ycotx = 2/(1 + sinx)`


The differential equation of all parabolas that have origin as vertex and y-axis as axis of symmetry is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×