English

F(x, y) = x2+y2x-y is a homogeneous function of degree 1.

Advertisements
Advertisements

Question

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

Options

  • True

  • False

MCQ
True or False
Advertisements

Solution

This statement is True.

Explanation:

Because f(λx, λy) = λ1f(x, y).

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Differential Equations - Solved Examples [Page 191]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 9 Differential Equations
Solved Examples | Q 23. (v) | Page 191

RELATED QUESTIONS

Show that the given differential equation is homogeneous and solve them.

(x2 + xy) dy = (x2 + y2) dx


Show that the given differential equation is homogeneous and solve them.

(x2 – y2) dx + 2xy dy = 0


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

(x + y) dy + (x – y) dx = 0; y = 1 when x = 1


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


Which of the following is a homogeneous differential equation?


\[xy \log\left( \frac{x}{y} \right) dx + \left\{ y^2 - x^2 \log\left( \frac{x}{y} \right) \right\} dy = 0\]

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

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

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

(x2 + 3xy + y2) dx − x2 dy = 0


(2x2 y + y3) dx + (xy2 − 3x3) dy = 0


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

Solve the following initial value problem:
 (x2 + y2) dx = 2xy dy, y (1) = 0


Solve the following initial value problem:
(xy − y2) dx − x2 dy = 0, y(1) = 1


Solve the following initial value problem:
\[\frac{dy}{dx} = \frac{y\left( x + 2y \right)}{x\left( 2x + y \right)}, y\left( 1 \right) = 2\]

 


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


Solve the following initial value problem:
\[\left\{ x \sin^2 \left( \frac{y}{x} \right) - y \right\}dx + x dy = 0, y\left( 1 \right) = \frac{\pi}{4}\]


Find the particular solution of the differential equation \[\left( x - y \right)\frac{dy}{dx} = x + 2y\], given that when x = 1, y = 0.


Which of the following is a homogeneous differential equation?


Solve the following differential equation:

x dx + 2y dx = 0, when x = 2, y = 1


Solve the following differential equation:

`x^2.  dy/dx = x^2 + xy + y^2`


Solve the following differential equation:

(x2 – y2)dx + 2xy dy = 0


Find the equation of a curve passing through `(1, pi/4)` if the slope of the tangent to the curve at any point P(x, y) is `y/x - cos^2  y/x`.


State the type of the differential equation for the equation. xdy – ydx = `sqrt(x^2 + y^2)  "d"x` and solve it


F(x, y) = `(ycos(y/x) + x)/(xcos(y/x))` is not a homogeneous function.


The differential equation y' = `y/(x + sqrt(xy))` has general solution given by:

(where C is a constant of integration)


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)

A function \[F(x,y)\] is homogeneous of degree \[n\] when which condition holds?


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


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


After putting \[y=vx\] in \[\frac{dy}{dx}=\frac{y\cos\left(\frac{y}{x}\right)+x}{x\cos\left(\frac{y}{x}\right)}\], which equation results?


Which separable equation follows from \[x\frac{dv}{dx}=\frac{1}{\cos v}\]?


Which final answer results after replacing \[v=\frac{y}{x}\] in \[\sin v=\ln|Cx|\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×