English

Solve the Differential Equation: X Dy - Y Dx = √ X 2 + Y 2 D X , Given that Y = 0 When X = 1. - Mathematics

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

xdy - ydx = `sqrt(x^2 + y^2)dx`

⇒ xdy = `[ y + sqrt(x^2+y^2)]dx`

`dy/dx = (y + sqrt(x^2+y^2))/x`       ...(1)

Let F (x,y) = `(y + sqrt(x^2+y^2))/x` 

∴ `"F"(lambdax,lambday) = (lambdax+sqrt((lambdax)^2+ (lambday)^2))/(lambdax) = (y + sqrt(x^2+y^2))/(x) = lambda^0 . "F"(x,y)`

Therefore, the given differential equation is a homogeneous equation.To solve it, we make the substitution as:

y = vx

⇒ `d/dx (y) = d/dx (vx)`
⇒ `dy/dx = v + x (dv)/(dx)`

Substituting the values of v and `dy/dx` in equation (1), we get:

`v + x (dv)/dx = (vx+sqrt(x^2 + (vx)^2))/x`

⇒ `v + x (dv)/dx = v + sqrt(1+v^2)`

⇒ `(dv)/sqrt(1+v^2) = dx/x`

Integrating both sides, we get:

`log |v + sqrt(1+v^2)| = log|x| + log "C"`

⇒ `log |y/x + sqrt(1+y^2/x^2)| = log|"C"x|`

⇒ `log|(y + sqrt(x^2+y^2))/x| = log|"C"x|`

⇒ `y + sqrt(x^2+y^2) = "C"x^2`

This is the required solution of the given differential equation.

shaalaa.com
  Is there an error in this question or solution?
2018-2019 (March) 65/1/1

RELATED QUESTIONS

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


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.

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


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

(x2 – y2) dx + 2xy dy = 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.

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


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

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


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

x2 dy + (xy + y2) 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.


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.


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

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

(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:
(xy − y2) dx − x2 dy = 0, y(1) = 1


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


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


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.


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


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`


State whether the following statement is True or False:   

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


The solution of the differential equation `(1 + e^(x/y)) dx + e^(x/y) (1 + x/y) dy` = 0 is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×