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

Determine the order and degree of the following differential equations. Solution D.E y = aex + be−x d2ydx2=1

Advertisements
Advertisements

प्रश्न

Determine the order and degree of the following differential equations.

Solution D.E
y = aex + be−x `(d^2y)/dx^2= 1`
बेरीज
Advertisements

उत्तर

y = aex + be−x  ......(1)

Differentiating w.r.t. x, we get

`dy/dx = ae^x - be^-x`

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

`(d^2y)/dx^2 = ae^x - be^-x`

∴ `(d^2y)/dx^2 = y` .....[From (i)]

∴ 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 2.5 | पृष्ठ १६२

संबंधित प्रश्‍न

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

Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]

 


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


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


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

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

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

Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.


Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\] 

 


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

y ex/y dx = (xex/y + y) dy


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

Solve the following initial value problem:-

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


Solve the following initial value problem:-

\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]


In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]


Find the equation of the curve passing through the point \[\left( 1, \frac{\pi}{4} \right)\]  and tangent at any point of which makes an angle tan−1  \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.


The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.


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


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


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

`y (1 + logx)dx/dy - x log x = 0`,

when x=e, y = e2.


Solve the following differential equation.

x2y dx − (x3 + y3) dy = 0


Solve the following differential equation.

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


Solve the differential equation:

`e^(dy/dx) = x`


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

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


Solve the following differential equation

`yx ("d"y)/("d"x)` = x2 + 2y2 


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 the following differential equation 

sec2 x tan y dx + sec2 y tan x dy = 0

Solution: sec2 x tan y dx + sec2 y tan x dy = 0

∴ `(sec^2x)/tanx  "d"x + square` = 0

Integrating, we get

`square + int (sec^2y)/tany  "d"y` = log c

Each of these integral is of the type

`int ("f'"(x))/("f"(x))  "d"x` = log |f(x)| + log c

∴ the general solution is

`square + log |tan y|` = log c

∴ log |tan x . tan y| = log c

`square`

This is the general solution.


Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.


Solution of `x("d"y)/("d"x) = y + x tan  y/x` is `sin(y/x)` = cx


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×