मराठी

Solve the Following Initial Value Problem: X D Y D X − Y + X Sin ( Y X ) = 0 , Y ( 2 ) = X - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

\[x\frac{dy}{dx} - y + x \sin\left( \frac{y}{x} \right) = 0, y(2) = \pi\]
It is a homogeneous equation . put y = vx
\[\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{ so,} v + x\frac{dv}{dx} = \frac{vx}{x} - \sin\left( \frac{vx}{x} \right)\]
\[x\frac{dv}{dx} = - \sin v\]
\[\frac{dv}{\sin v} = - \frac{dx}{x}\]
\[\text{ cosec }(v)dv = - \frac{dx}{x}\]
Integraing both sides we get,
\[\log(\text{cosec }(v) - cot(v)) = - \log x + \log c\]
\[log\left( \text{cosec }\left( \frac{y}{x} \right) - cot\left( \frac{y}{x} \right) \right) = - log x + log c\]
\[\text{Putting the values }x = 2\text{ and }y = \pi \]
\[log\left(\text{cosec }\left( \frac{\pi}{2} \right) - cot\left( \frac{\pi}{2} \right) \right) = - log 2 + log c\]
\[c = 0\]
\[log\left( \text{cosec }\left( \frac{y}{x} \right) - cot\left( \frac{y}{x} \right) \right) = - log x\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 22: Differential Equations - Exercise 22.09 [पृष्ठ ८४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Exercise 22.09 | Q 36.9 | पृष्ठ ८४

संबंधित प्रश्‍न

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


Solve the differential equation :

`y+x dy/dx=x−y dy/dx`


 

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.

(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`


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


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


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


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

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

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

(x2 + 3xy + y2) dx − x2 dy = 0


(2x2 y + y3) dx + (xy2 − 3x3) dy = 0


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

 


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


Solve the following differential equation : \[\left[ y - x  \cos\left( \frac{y}{x} \right) \right]dy + \left[ y  \cos\left( \frac{y}{x} \right) - 2x  \sin\left( \frac{y}{x} \right) \right]dx = 0\] .


Solve the following differential equation:

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


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:

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


Solve the following differential equation:

`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`


Solve the following differential equation:

(x2 – y2)dx + 2xy dy = 0


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


Solve : `x^2 "dy"/"dx"` = x2 + xy + y2.


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 ______.


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

(where C is a constant of integration)


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)

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×