मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Solve the following differential equation: x⋅dydx-y+x⋅sin(yx)=0

Advertisements
Advertisements

प्रश्न

Solve the following differential equation:

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

बेरीज
Advertisements

उत्तर १

`"x" "dy"/"dx" - "y" + "x sin"("y"/"x") = 0`   ...(1)

Put y = vx

∴ `"dy"/"dx" = "v + x" "dv"/"dx" and "y"/"x" = "v"`

∴ equation (1) becomes,

`x("v + x""dv"/"dx") - "vx + x sin v" = 0`

∴ `"vx" + "x"^2 "dv"/"dx" - "vx" + "x sin v" = 0`

∴ `"x"^2 "dv"/"dx" + "x sin v" = 0`

∴ `1/"sin v" "dv" + 1/"x" "dx" = 0`

Integrating, we get

∴ `int "cosec v dv" + int1/"x" "dx" = "c"_1`

∴ `log |"cosec v - cot v"| + log |"x"| = log "c"`, where c1 = log c

∴ `log |"x" ("cosec v" - "cot v")| = log "c"`

∴ `"x"(1/(sin"v") - (cos "v")/(sin"v")) = "c"`

∴ x(1 - cos v) = c sin v

∴ `"x"[1 - cos("y"/"x")] = "c sin"("y"/"x")`

This is the general solution.

shaalaa.com

उत्तर २

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

Put y = vx

⇒ `dy/dx = v + x (dv)/dx`

∴ The given equation becomes, 

`x (v + x (dv)/dx) - vx + x sin v = 0`

∴ `vx + x^2 (dv)/dx - vx + x sin v = 0`

∴ `x^2 (dv)/dx = -x sin v`

∴ `(dv)/(sin v) = - dx/x`    ...(variable separable form)

Integrating on both sides, we get,

`int cosec  v  dv = -int 1/x dx`

∴ log (cosec v − cot v) = − log x + log c

∴ log (cosec v − cot v) = `log (c/x)`

∴ cosec v − cot v = `c/x`

∴ `x [cosec (y/x) - cot (y/x)] = c` is the required solution.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Differential Equations - Exercise 6.4 [पृष्ठ २०३]

संबंधित प्रश्‍न

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


Solve the differential equation :

`y+x dy/dx=x−y dy/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:

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


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.


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

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

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

(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:
 (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:
(xy − y2) dx − x2 dy = 0, y(1) = 1


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


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


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:

`"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:

y2 dx + (xy + x2)dy = 0


Solve the following differential equation:

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


Solve the following differential equation:

`"y"^2 - "x"^2 "dy"/"dx" = "xy""dy"/"dx"`


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


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) = `(sqrt(x^2 + y^2) + y)/x` is a homogeneous function of degree ______.


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


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 substitutions are chosen for a homogeneous differential equation?


Writing \[x\cos\left(\frac{y}{x}\right)\frac{dy}{dx}=y\cos\left(\frac{y}{x}\right)+x\] in standard form gives which expression?


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?


From \[v+x\frac{dv}{dx}=\frac{v\cos v+1}{\cos v}\], what is \[x\frac{dv}{dx}\]?


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×