मराठी

A homogeneous differential equation of the from dxdy=h(xy) can be solved by making the substitution. - Mathematics

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • y = vx

  • v = yx

  • x = vy

  • x = v

MCQ
Advertisements

उत्तर

x = vy

Explanation:

By substituting x = vy, where `v = x/y`, the differential equation `dx/dy = h (x/y)` is transformed into a separable from, making it easier to solve.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Equations - Exercise 9.5 [पृष्ठ ४०६]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 9 Differential Equations
Exercise 9.5 | Q 16 | पृष्ठ ४०६

संबंधित प्रश्‍न

Solve the differential equation (x2 + y2)dx- 2xydy = 0


Solve the differential equation :

`y+x dy/dx=x−y dy/dx`


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.

`y' = (x + y)/x`


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

(x – y) dy – (x + y) dx = 0


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

`x^2 dy/dx = x^2 - 2y^2 + xy`


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

`x dy/dx - y +  x sin (y/x) = 0`


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

`[xsin^2(y/x - y)] dx + x  dy = 0; y = pi/4 "when"  x = 1`


Which of the following is a homogeneous differential equation?


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


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

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

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

(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:
\[\frac{dy}{dx} - \frac{y}{x} + cosec\frac{y}{x} = 0, y\left( 1 \right) = 0\]


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, \[y = \frac{\pi}{4}\]


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


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


Solve the differential equation: x dy - y dx = `sqrt(x^2 + y^2)dx,` given that y = 0 when x = 1.


Solve the following differential equation:

`"x" sin ("y"/"x") "dy" = ["y" sin ("y"/"x") - "x"] "dx"`


Solve the following differential equation:

`(1 + 2"e"^("x"/"y")) + 2"e"^("x"/"y")(1 - "x"/"y") "dy"/"dx" = 0`


Solve the following differential equation:

`(1 + "e"^("x"/"y"))"dx" + "e"^("x"/"y")(1 - "x"/"y")"dy" = 0`


Solve the following differential equation:

`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`


Solve the following differential equation:

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


Solve the following differential equation:

(x2 + 3xy + y2)dx - x2 dy = 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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×