हिंदी

Solve the Following Initial Value Problem: (Xy − Y2) Dx − X2 Dy = 0, Y(1) = 1

Advertisements
Advertisements

प्रश्न

Solve the following initial value problem:
(xy − y2) dx − x2 dy = 0, y(1) = 1

Advertisements

उत्तर

(xy − y2) dx − x2 dy = 0, y(1) = 1
This is an homogenous equation, put y = vx
\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\left( xy - y^2 \right) = x^2 \left( \frac{dy}{dx} \right)\]
\[\left( v x^2 - v^2 x^2 \right) = x^2 \left( v + x\frac{dv}{dx} \right)\]
\[v x^2 \left( 1 - v \right) = x^2 \left( v + x\frac{dv}{dx} \right)\]
\[v\left( 1 - v \right) = v + x\frac{dv}{dx}\]
\[v - v^2 = v + x\frac{dv}{dx}\]
\[ - v^2 = x\frac{dv}{dx}\]
\[ - \frac{1}{x}dx = \frac{1}{v^2}dv\]
On integrating both sides we get,
\[- \int\frac{1}{x}dx = \int\frac{1}{v^2}dv\]
\[ - \log_e x = \frac{v^{- 2 + 1}}{- 2 + 1} + c\]
\[ - \log_e x = \frac{v^{- 1}}{- 1} + c\]
\[ - \log_e x = - \frac{1}{v} + c\]
\[ - \log_e x = - \frac{1}{v} + c\]
\[\frac{x}{y} - \log_e x = c\]
\[\text{ As }y\left( 1 \right) = 1\]
\[\frac{1}{1} - \log_e 1 = c\]
\[ \Rightarrow c = 1\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 21: Differential Equations - Exercise 22.09 [पृष्ठ ८४]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 21 Differential Equations
Exercise 22.09 | Q 36.4 | पृष्ठ ८४

संबंधित प्रश्न

 

Show that the differential  equation `2xydy/dx=x^2+3y^2`  is homogeneous and solve it.

 

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

`y' = (x + y)/x`


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/dx - y +  x sin (y/x) = 0`


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

`y  dx + x log(y/x)dy - 2x  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`


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

`2xy + y^2 - 2x^2  dy/dx = 0; y = 2`   when x  = 1


Find the particular solution of the differential equation `(x - y) dy/dx = (x + 2y)` given that y = 0 when x = 1.


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

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

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

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

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

(x2 + 3xy + y2) dx − x2 dy = 0


(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:
\[\frac{dy}{dx} - \frac{y}{x} + cosec\frac{y}{x} = 0, y\left( 1 \right) = 0\]


Show that the family of curves for which \[\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\], is given by \[x^2 - y^2 = Cx\]


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


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" sin ("y"/"x") "dy" = ["y" sin ("y"/"x") - "x"] "dx"`


Solve the following differential equation:

y2 dx + (xy + x2)dy = 0


Solve the following differential equation:

`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 0`


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:

(x2 – y2)dx + 2xy dy = 0


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


F(x, y) = `(x^2 + y^2)/(x - y)` is a homogeneous function of degree 1.


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 ______.


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 equation `dy/dx = (3x − 4y − 2)/(3x − 4y − 3)` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×