मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

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

Advertisements
Advertisements

प्रश्न

Solve the following differential equation.

(x2 − y2 ) dx + 2xy dy = 0

बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Differential Equation and Applications - Exercise 8.4 [पृष्ठ १६७]

APPEARS IN

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

संबंधित प्रश्‍न

\[\sqrt[3]{\frac{d^2 y}{d x^2}} = \sqrt{\frac{dy}{dx}}\]

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


Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]

 


Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]

Function y = log x


Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x


Solve the differential equation \[\frac{dy}{dx} = e^{x + y} + x^2 e^y\].

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

y (1 + ex) dy = (y + 1) ex dx


Solve the following differential equation: 
(xy2 + 2x) dx + (x2 y + 2y) dy = 0


\[\frac{dy}{dx} = y \tan 2x, y\left( 0 \right) = 2\] 

\[\frac{dy}{dx} = y \sin 2x, y\left( 0 \right) = 1\]

\[\frac{dy}{dx} = 2 e^{2x} y^2 , y\left( 0 \right) = - 1\]

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

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

Solve the following initial value problem:-

\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]


Solve the following initial value problem:-

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


The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?


The tangent at any point (x, y) of a curve makes an angle tan−1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).


Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.


The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by


Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?


In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-

y = ex + 1            y'' − y' = 0


In each of the following examples, verify that the given function is a solution of the corresponding differential equation.

Solution D.E.
y = ex  `dy/ dx= y`

Solve the following differential equation.

`dy/dx + y = e ^-x`


Solve the following differential equation y log y = `(log  y - x) ("d"y)/("d"x)`


Choose the correct alternative:

Solution of the equation `x("d"y)/("d"x)` = y log y is


Solve: `("d"y)/("d"x) = cos(x + y) + sin(x + y)`. [Hint: Substitute x + y = z]


If `y = log_2 log_2(x)` then `(dy)/(dx)` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×