हिंदी

Solve the following differential equation. (x2 − y2 ) dx + 2xy dy = 0

Advertisements
Advertisements

प्रश्न

Solve the following differential equation.

(x2 − y2 ) dx + 2xy dy = 0

योग
Advertisements

उत्तर

(x2 − y2 ) dx + 2xy dy = 0

∴ 2xy dy = (y2 - x2) dx

∴ `dy/dx = (y^2 - x^2)/(2xy) ......(i)`

Put y = tx  ...(ii)

Differentiating w.r.t. x, we get

`dy/dx = t +x dt/dx  ...(iii)`

Substituting (ii) and (iii) in (i), we get

`t + x dt/dx = (t^2 x^2-x^2)/(2tx^2)`

∴ `x dt/dx = (t^2 - 1)/(2t )- t = (-(1+t^2))/(2t)`

∴ `2t/(1+t^2) dt = - dx/x`

Integrating on both sides, we get

`int 2t/(1+t^2) dt = - int dx/x`

∴ log |1 + t2| = -log |x| + log |c|

∴`log | 1+y^2/x^2| = log |c/x|`

∴ `(x^2 + y^2)/x^2 = c/x`

∴  x2 + y2 = cx

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Differential Equation and Applications - Exercise 8.4 [पृष्ठ १६७]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 8 Differential Equation and Applications
Exercise 8.4 | Q 1.5 | पृष्ठ १६७

संबंधित प्रश्न

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

Verify that \[y = ce^{tan^{- 1}} x\]  is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]


Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2


\[\frac{dy}{dx} = \frac{1 - \cos 2y}{1 + \cos 2y}\]

\[\frac{dy}{dx} = e^{x + y} + e^y x^3\]

\[x\sqrt{1 - y^2} dx + y\sqrt{1 - x^2} dy = 0\]

\[\frac{dy}{dx} = 1 - x + y - xy\]

\[2x\frac{dy}{dx} = 5y, y\left( 1 \right) = 1\]

\[\frac{dy}{dx} = 1 + x + y^2 + x y^2\] when y = 0, x = 0

In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).


\[\cos^2 \left( x - 2y \right) = 1 - 2\frac{dy}{dx}\]

(x + y) (dx − dy) = dx + dy


3x2 dy = (3xy + y2) dx


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

 

Solve the following initial value problem:-

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


At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.


The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.


Define a differential equation.


The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when


Find the differential equation whose general solution is

x3 + y3 = 35ax.


The solution of `dy/dx + x^2/y^2 = 0` is ______


State whether the following is True or False:

The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.


y dx – x dy + log x dx = 0


Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0


Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`


The solution of differential equation `x^2 ("d"^2y)/("d"x^2)` = 1 is ______


The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`


Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.


Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.


Solve the differential equation

`x + y dy/dx` = x2 + y2


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×