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

Find the differential equation whose general solution is x3 + y3 = 35ax.

Advertisements
Advertisements

प्रश्न

Find the differential equation whose general solution is

x3 + y3 = 35ax.

बेरीज
Advertisements

उत्तर

x3 + y3 = 35ax ...(i)

Differentiating w.r.t. x, we get

`3x^3 + 3y^3 dy/dx = 35a` ...(ii)

Substituting (ii) in (i), we get

`x^3 + y^3 = (3x^2 + 3y^2 dy/dx)x`

∴ `x^3 + y^3 = 3x^3 + 3x*y^2 dy/dx`

∴ `2x^3 - y^3 +3xy^2dy/dx =0`, which is the required differential equation.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Differential Equation and Applications - Exercise 8.2 [पृष्ठ १६३]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 8 Differential Equation and Applications
Exercise 8.2 | Q 4 | पृष्ठ १६३

संबंधित प्रश्‍न

Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]


Verify that y = cx + 2c2 is a solution of the differential equation 

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

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

\[\left( x^3 + x^2 + x + 1 \right)\frac{dy}{dx} = 2 x^2 + x\]

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

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

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

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


\[2x\frac{dy}{dx} = 3y, y\left( 1 \right) = 2\]

\[\frac{dy}{dx} = 2 e^x y^3 , y\left( 0 \right) = \frac{1}{2}\]

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

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

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

(y2 − 2xy) dx = (x2 − 2xy) dy


Solve the following initial value problem:-

\[y' + y = e^x , y\left( 0 \right) = \frac{1}{2}\]


Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]


Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\]  = x (x + 1) and passing through (1, 0).


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.


Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\]  are rectangular hyperbola.


The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by


The differential equation satisfied by ax2 + by2 = 1 is


The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting


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


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

Solution D.E.
xy = log y + k y' (1 - xy) = y2

Solve the following differential equation.

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


For each of the following differential equations find the particular solution.

(x − y2 x) dx − (y + x2 y) dy = 0, when x = 2, y = 0


Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`


Solve: `("d"y)/("d"x) + 2/xy` = x2 


`d/(dx)(tan^-1  (sqrt(1 + x^2) - 1)/x)` is equal to:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×