हिंदी

Show that the given differential equation is homogeneous and solve them. (1+exy)dx+exy(1-xy)dy=0

Advertisements
Advertisements

प्रश्न

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

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

योग
Advertisements

उत्तर

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

`=> dx/dy = (e^(x/y)(x/y - 1))/(1 + e^(x/y)) = g(x/y)`     say   ...(i)

∵ The right side of the equation is in the form of `g (x/y)`, so it is a homogeneous differential equation of zero degrees.

∴Putting  x = vy

`dx/dy = v + y (dv)/dy`    ...(from equation (i))

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

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

`=> (ve^v - e^v - v - ve^v)/(1 + e^v)`

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

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

⇒ log |ev + v| = - log |y| + C1

⇒ log |(ev + v)y| = C1

⇒ |(ev + v) y| = eC1

⇒ (ev + v)y = ± eC1 = C  (say)

⇒ `(e^(x/y) + x/y) y = C`

⇒ `y  e^(x/y) + x = C`

which is the required general solution of the given differential equation.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Equations - Exercise 9.5 [पृष्ठ ४०६]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 9 Differential Equations
Exercise 9.5 | Q 10 | पृष्ठ ४०६

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.


Solve the differential equation :

`y+x dy/dx=x−y dy/dx`


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.

(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.

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


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

x2 dy + (xy + y2) dx = 0; y = 1 when x = 1


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`


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

`dy/dx -  y/x + cosec (y/x) = 0; y = 0` 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 the differential equation (x3 – 3xy2)dx = (y3 – 3x2y)dy, where C is parameter


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

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

\[\left( x^2 + y^2 \right)\frac{dy}{dx} = 8 x^2 - 3xy + 2 y^2\]

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

Solve the following initial value problem:
 (x2 + y2) dx = 2xy dy, y (1) = 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


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


Solve the following differential equation:

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


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`


Solve the following differential equation:

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


Solve the following differential equation:

(x2 + 3xy + y2)dx - x2 dy = 0


Solve the following differential equation:

(x2 – y2)dx + 2xy dy = 0


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


A function \[F(x,y)\] is homogeneous of degree \[n\] when which condition holds?


If \[F(\lambda x,\lambda y)=F(x,y)\] for any non-zero constant \[\lambda\], what is the degree of \[F(x,y)\]?


After using \[y=vx\] and writing the right-hand side as \[g(v)\], which separable form is obtained?


Which dependence is checked on the right-hand side of a homogeneous differential equation?


What form should the equation be reduced to before integration?


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×