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

Determine the order and degree of the following differential equations. Solution D.E. ax2 + by2 = 5 xyd2ydx2+x(dydx)2=ydydx

Advertisements
Advertisements

प्रश्न

Determine the order and degree of the following differential equations.

Solution D.E.
ax2 + by2 = 5 `xy(d^2y)/dx^2+ x(dy/dx)^2 = y dy/dx`
बेरीज
Advertisements

उत्तर

ax 2 + by 2 = 5

Differentiating w.r.t. x, we get

`2ax +2by dy/dx = 0` ....(i)

Again, differentiating w.r.t. x, we get

`2a + 2b(dy/dx)^2 + 2by ((d^2y)/dx^2) = 0` ........(ii)

From (i), we get

`a = - (by)/x(dy/dx)`

Substituting the value of a in (ii), we get

`- 2(by)/x(dy/dx) + 2b(dy/dx)^2 + 2by((d^2y)/dx^2) = 0`

∴`- y/x(dy/dx) + (dy/dx)^2 + y((d^2y)/dx^2) = 0`

∴`- y(dy/dx) + x(dy/dx)^2 + xy((d^2y)/dx^2) = 0`

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

∴ Given function is a solution of the given differential equation.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Differential Equation and Applications - Exercise 8.1 [पृष्ठ १६२]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 8 Differential Equation and Applications
Exercise 8.1 | Q 2.6 | पृष्ठ १६२

संबंधित प्रश्‍न

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


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

C' (x) = 2 + 0.15 x ; C(0) = 100


(1 − x2) dy + xy dx = xy2 dx


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

\[2\left( y + 3 \right) - xy\frac{dy}{dx} = 0\], y(1) = −2

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

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

(x + y) (dx − dy) = dx + dy


\[\frac{dy}{dx} + 1 = e^{x + y}\]

x2 dy + y (x + y) dx = 0


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


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

 

Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]


Solve the following initial value problem:-

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


The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.


The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).


Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.


Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]


Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.


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

Solution D.E.
y = xn `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0`

Solve the following differential equation.

`y^3 - dy/dx = x dy/dx`


For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1


Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0


Solve the following differential equation.

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


Solve the following differential equation.

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


Solve the differential equation:

`e^(dy/dx) = x`


Solve the following differential equation

`y log y ("d"x)/("d"y) + x` = log y


Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×