English

Show that the differential equation 2y^(x/y) dx + (y − 2x e^(x/y)) dy = 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1. - Mathematics

Advertisements
Advertisements

Question

Show that the differential equation 2yx/y dx + (y − 2x ex/y) dy = 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1.

Advertisements

Solution

The given differential equation can be written as

`dx/dy=(2xe^(x/y)-y)/(2ye^(x/y)) .....................(1)`

`Let F(x,y)=(2xe^(x/y)-y)/(2ye^(x/y))`

then ` F(lambdax,lambday)=(lambda(2xe^(x/y)-y))/(lambda(2ye^(x/y)))=lambda^@[F(x,y)]`

Thus, F (x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.

For solving, let us substitute x=vy     ..................(2)

Differentiating equation (2) with respect to y, we get

`dx/dy=v+y(dv)/(dy)`

Substituting the value of x and  `dx/dy ` in equation (1), we get

`v+y(dv)/(dy)=(2ve^v-1)/(2e^v)`

`or y(dv)/(dy)=(2ve^v-1)/(2e^v)-v`

`or y(dv)/(dy)=-1/(2e^v)`

`or 2e^vdv=(-dy)/y`

`or int 2e^vdv=-intdy/y`

`or 2e^v=-log|y|+C`

Replacing v by x/y , we get

`2e^(x/y)+log|y|=c......(3)`

Substituting x = 0 and y = 1in equation (3), we get

`2e^0+log|1|=c =>c=2`

Substituting the value of C in equation (3), we get

`2e^(x/y)+log|y|=2` .which is the particular solution of the given differential equation.

 

shaalaa.com
  Is there an error in this question or solution?
2012-2013 (March) Delhi Set 1

RELATED QUESTIONS

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.

(x – y) dy – (x + y) dx = 0


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

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


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

`y  dx + x log(y/x)dy - 2x  dy = 0`


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

`(1+e^(x/y))dx + e^(x/y) (1 - x/y)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


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

(x2 − 2xy) dy + (x2 − 3xy + 2y2) dx = 0


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

(x2 + 3xy + y2) dx − x2 dy = 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:
x (x2 + 3y2) dx + y (y2 + 3x2) 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.


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:

`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 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`


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


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


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


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

(where C is a constant of integration)


Find the general solution of the differential equation:

(xy – x2) dy = y2 dx


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×