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

Xy dydx =x2+2y2 - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

`xy dy/dx  = x^2 + 2y^2`

बेरीज
Advertisements

उत्तर

`xy  dy/dx  = x^2 + 2y^2`

∴ `dy/dx = x^2 + (2y^2)/ (xy)`  …(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 = (x^2 + 2t^2 x^2)/ (x(tx))`

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

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

∴ `t/(1+t^2) dt = 1/x dx`

Integrating on both sides, we get

`1/2 int (2t)/(1+t^2)  dt = int dx/x`

∴ `1/2 log|1 + t^2| = log |x| + log |c|`

∴ log |1 + t2 | = 2 log |x| + 2log |c|

= log |x2 | + log |c2|

∴ log |1 + t2 | = log |c2 x2|

∴ 1 + t2 = c2x2

∴ `1 + y^2/x^2 = c^2x^2`

∴ x2 + y2 = c2 x4

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Differential Equation and Applications - Miscellaneous Exercise 8 [पृष्ठ १७३]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 8 Differential Equation and Applications
Miscellaneous Exercise 8 | Q 4.13 | पृष्ठ १७३

संबंधित प्रश्‍न

Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]


Verify that y = 4 sin 3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 9y = 0\]


For the following differential equation verify that the accompanying function is a solution:

Differential equation Function
\[x\frac{dy}{dx} + y = y^2\]
\[y = \frac{a}{x + a}\]

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


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

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

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

(ey + 1) cos x dx + ey sin x dy = 0


xy dy = (y − 1) (x + 1) dx


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

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

(1 − x2) dy + xy dx = xy2 dx


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

\[\frac{dy}{dx} = \frac{x}{2y + x}\]

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

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:-

\[\frac{dy}{dx} + 2y \tan x = \sin x; y = 0\text{ when }x = \frac{\pi}{3}\]


The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.


Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.


The differential equation `y dy/dx + x = 0` represents family of ______.


Form the differential equation from the relation x2 + 4y2 = 4b2


Solve the differential equation:

`e^(dy/dx) = x`


Solve

`dy/dx + 2/ x y = x^2`


Select and write the correct alternative from the given option for the question 

Differential equation of the function c + 4yx = 0 is


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


Solve the differential equation xdx + 2ydy = 0


For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0


A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution


Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×