मराठी

Which of the following is not a homogeneous function of x and y. - Mathematics

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प्रश्न

Which of the following is not a homogeneous function of x and y.

पर्याय

  • x2 + 2xy

  • 2x – y

  • `cos^2 (y/x) + y/x`

  • sinx – cosy

MCQ
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उत्तर

sinx – cosy

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पाठ 9: Differential Equations - Solved Examples [पृष्ठ १८८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 9 Differential Equations
Solved Examples | Q 19 | पृष्ठ १८८

संबंधित प्रश्‍न

 

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.


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

`{xcos(y/x) + ysin(y/x)}ydx = {ysin (y/x) -  xcos(y/x)}xdy`


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

`y  dx + x log(y/x)dy - 2x  dy = 0`


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?


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

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

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

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

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


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

 


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


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


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


A homogeneous differential equation of the form \[\frac{dx}{dy} = h\left( \frac{x}{y} \right)\] can be solved by making the substitution


Which of the following is a homogeneous differential equation?


Solve the following differential equation : \[\left[ y - x  \cos\left( \frac{y}{x} \right) \right]dy + \left[ y  \cos\left( \frac{y}{x} \right) - 2x  \sin\left( \frac{y}{x} \right) \right]dx = 0\] .


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


Solve the following differential equation:

y2 dx + (xy + x2)dy = 0


Solve the following differential equation:

`x * dy/dx - y + x * sin(y/x) = 0`


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


Solcve: `x ("d"y)/("d"x) = y(log y – log x + 1)`


Let the solution curve of the differential equation `x (dy)/(dx) - y = sqrt(y^2 + 16x^2)`, y(1) = 3 be y = y(x). Then y(2) is equal to ______.


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)

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