English

Show that Y = Aebx is a Solution of the Differential Equation D 2 Y D X 2 = 1 Y ( D Y D X ) 2

Advertisements
Advertisements

Question

Show that y = AeBx is a solution of the differential equation

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

Solution

We have, \[y = A e^{Bx}  ................(1)\]
Differentiating both sides of (1) with respect to x, we get
\[\frac{dy}{dx} = AB e^{Bx}   ................(2)\]
Differentiating both sides of (2) with respect to x, we get
\[\frac{d^2 y}{d x^2} = A B^2 e^{Bx} \]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{\left( AB e^{Bx} \right)^2}{\left( A e^{Bx} \right)}\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{1}{y} \left( \frac{dy}{dx} \right)^2 ...........\left[\text{Using }\left( 1 \right)\text{ and }\left( 2 \right) \right]\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{1}{y} \left( \frac{dy}{dx} \right)^2\]

Hence, the given function is the solution to the given differential equation.

shaalaa.com
  Is there an error in this question or solution?
Chapter 21: Differential Equations - Exercise 22.03 [Page 25]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.03 | Q 6 | Page 25

RELATED QUESTIONS

Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].


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


\[\sin^4 x\frac{dy}{dx} = \cos x\]

\[\frac{dy}{dx} = x e^x - \frac{5}{2} + \cos^2 x\]

xy (y + 1) dy = (x2 + 1) dx


\[5\frac{dy}{dx} = e^x y^4\]

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

(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0


(y2 + 1) dx − (x2 + 1) dy = 0


Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]


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

\[\frac{dr}{dt} = - rt, r\left( 0 \right) = r_0\]

Find the particular solution of the differential equation
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.


\[\frac{dy}{dx} = \frac{\left( x - y \right) + 3}{2\left( x - y \right) + 5}\]

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

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

Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]


Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.

 


The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?


If the marginal cost of manufacturing a certain item is given by C' (x) = \[\frac{dC}{dx}\] = 2 + 0.15 x. Find the total cost function C (x), given that C (0) = 100.

 

Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]


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.


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 differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is


The solution of the differential equation y1 y3 = y22 is


The differential equation satisfied by ax2 + by2 = 1 is


What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?


Find the differential equation whose general solution is

x3 + y3 = 35ax.


Solve the following differential equation.

`dy/dx + 2xy = x`


Choose the correct alternative.

The integrating factor of `dy/dx -  y = e^x `is ex, then its solution is


A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.


`xy dy/dx  = x^2 + 2y^2`


y dx – x dy + log x dx = 0


Solve: `("d"y)/("d"x) + 2/xy` = x2 


For the differential equation, find the particular solution

`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0


Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0


Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.


Solve: ydx – xdy = x2ydx.


Solve the differential equation

`y (dy)/(dx) + x` = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×