मराठी

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. - Mathematics

Advertisements
Advertisements

प्रश्न

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.

बेरीज
Advertisements

उत्तर

`2ye^(x/y)dx+(y-2xe^(x/y))dy=0`

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

Given differential equation is a homogeneous differential equation.

∴ Put x = vy

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

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

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

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

`=>2e^vdv=-1/ydy`

Integrating on both the sides

`=>2inte^vdv=-int1/ydy`

`=>2e^v=-log|y|+logC`

`=>2e^v=log|c/y|`

`=>2e^(x/y)=log|c/y|`

Given that at x = 0, y = 1

`2e^0= log|c/1|`

⇒ C = e2

`:.2e^(x/y)=log""e^2/y`

`=>logy=-2e^(x/y)+2`

`=>y=e^2-2e^(x/y)`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2015-2016 (March) All India Set 1 N

संबंधित प्रश्‍न

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.


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`


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.


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


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

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

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


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

\[x\frac{dy}{dx} - y + x \sin\left( \frac{y}{x} \right) = 0\]

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

 


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


Which of the following is a homogeneous differential equation?


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


Solve the following differential equation:

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


Solve the following differential equation:

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


Solve the following differential equation:

(9x + 5y) dy + (15x + 11y)dx = 0


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


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


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


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 ______.


Find the general solution of the differential equation:

(xy – x2) dy = y2 dx


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×