English

For the Following Differential Equation Verify that the Accompanying Function is a Solution: Differential Equation Function X + Y D Y D X = 0 Y = ± √ a 2 − X 2 - Mathematics

Advertisements
Advertisements

Question

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

Differential equation Function
\[x + y\frac{dy}{dx} = 0\]
\[y = \pm \sqrt{a^2 - x^2}\]
Advertisements

Solution

We have,
\[y = \pm \sqrt{a^2 - x^2}\]
\[ \Rightarrow y^2 = a^2 - x^2 . . . . . \left( 1 \right)\]
Given differential equation:
\[x + y\frac{dy}{dx} = 0\]
Differentiating both sides of (1) with respect to x, we get
\[2y \frac{dy}{dx} = - 2x\]
\[ \Rightarrow y \frac{dy}{dx} = - x\]
\[ \Rightarrow x + y \frac{dy}{dx} = 0\]
Hence, the given function is the solution to the given differential equation.

shaalaa.com
  Is there an error in this question or solution?
Chapter 22: Differential Equations - Exercise 22.03 [Page 25]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 22 Differential Equations
Exercise 22.03 | Q 21.2 | Page 25

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

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

 


Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.


Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\]  satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]


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

Differential equation Function
\[y = \left( \frac{dy}{dx} \right)^2\]
\[y = \frac{1}{4} \left( x \pm a \right)^2\]

Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex


\[\frac{dy}{dx} = \tan^{- 1} x\]


\[\frac{dy}{dx} = \log x\]

\[\frac{dy}{dx} = \cos^3 x \sin^2 x + x\sqrt{2x + 1}\]

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

(1 + x2) dy = xy dx


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

y (1 + ex) dy = (y + 1) ex dx


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

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.


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


In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).


In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.


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

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

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

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

(x + 2y) dx − (2x − y) dy = 0


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

Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]


A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.


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


Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].


The price of six different commodities for years 2009 and year 2011 are as follows: 

Commodities A B C D E F

Price in 2009 (₹)

35 80 25 30 80 x
Price in 2011 (₹) 50 y 45 70 120 105

The Index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x andy if the total price in 2009 is ₹ 360.


Choose the correct option from the given alternatives:

The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is


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

Solution D.E.
y = ex  `dy/ dx= y`

Determine the order and degree of the following differential equations.

Solution D.E.
y = 1 − logx `x^2(d^2y)/dx^2 = 1`

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.


Choose the correct alternative:

Solution of the equation `x("d"y)/("d"x)` = y log y is


The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`


Solve the following differential equation `("d"y)/("d"x)` = x2y + y


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×