English

State whether the following statement is True or False: A homogeneous differential equation is solved by substituting y = vx and integrating it

Advertisements
Advertisements

Question

State whether the following statement is True or False:   

A homogeneous differential equation is solved by substituting y = vx and integrating it

Options

  • True

  • False

MCQ
True or False
Advertisements

Solution

True

shaalaa.com
  Is there an error in this question or solution?
Chapter 1.8: Differential Equation and Applications - Q.3

APPEARS IN

RELATED QUESTIONS

Solve the differential equation :

`y+x dy/dx=x−y dy/dx`


 

Show that the differential  equation `2xydy/dx=x^2+3y^2`  is homogeneous and solve it.

 

Find the particular solution of the differential equation:

2y ex/y dx + (y - 2x ex/y) dy = 0 given that x = 0 when y = 1.


For the differential equation find a particular solution satisfying the given condition:

`2xy + y^2 - 2x^2  dy/dx = 0; y = 2`   when x  = 1


A homogeneous differential equation of the from `dx/dy = h (x/y)` can be solved by making the substitution.


Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation  (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter.


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


Prove that x2 – y2 = c(x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2)dx = (y3 – 3x2y)dy, where C is parameter


\[\frac{y}{x}\cos\left( \frac{y}{x} \right) dx - \left\{ \frac{x}{y}\sin\left( \frac{y}{x} \right) + \cos\left( \frac{y}{x} \right) \right\} dy = 0\]

\[x\frac{dy}{dx} - y = 2\sqrt{y^2 - x^2}\]

(x2 + 3xy + y2) dx − x2 dy = 0


Solve the following initial value problem:
\[x e^{y/x} - y + x\frac{dy}{dx} = 0, y\left( e \right) = 0\]


Solve the following initial value problem:
\[\frac{dy}{dx} - \frac{y}{x} + cosec\frac{y}{x} = 0, y\left( 1 \right) = 0\]


Solve the following initial value problem:
(y4 − 2x3 y) dx + (x4 − 2xy3) dy = 0, y (1) = 1


Solve the following initial value problem:
x (x2 + 3y2) dx + y (y2 + 3x2) dy = 0, y (1) = 1


Find the particular solution of the differential equation x cos\[\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\], given that when x = 1, \[y = \frac{\pi}{4}\]


Show that the family of curves for which \[\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\], is given by \[x^2 - y^2 = Cx\]


Solve the differential equation:  ` (dy)/(dx) = (x + y )/ (x - y )`


F(x, y) = `(x^2 + y^2)/(x - y)` is a homogeneous function of degree 1.


If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation, 2x2dy = (2xy + y2)dx, then `f(1/2)` is equal to ______.


Read the following passage:

An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form `dy/dx` = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogeneous function of degree n if F(λx, λy) = λn F(x, y).

To solve a homogeneous differential equation of the type `dy/dx` = F(x, y) = `g(y/x)`, we make the substitution y = vx and then separate the variables.

Based on the above, answer the following questions:

  1. Show that (x2 – y2) dx + 2xy dy = 0 is a differential equation of the type `dy/dx = g(y/x)`. (2)
  2. Solve the above equation to find its general solution. (2)

The solution of the differential equation y2 dx + (x2 − xy + y2)dy = 0 is ______.


After using \[y=vx\] and writing the right-hand side as \[g(v)\], which separable form is obtained?


For the substitution \[x=vy\], which differentiated form is correct?


Which dependence is checked on the right-hand side of a homogeneous differential equation?


What replacement obtains the final answer after using a homogeneous substitution?


For \[F(x,y)=\frac{y\cos\left(\frac{y}{x}\right)+x}{x\cos\left(\frac{y}{x}\right)}\], what is \[F(\lambda x,\lambda y)\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×