English

Show that the Differential Equation of Which Y = 2(X2 − 1) + C E − X 2 is a Solution, is D Y D X + 2 X Y = 4 X 3 - Mathematics

Advertisements
Advertisements

Question

Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]

Solution

The given equation is \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\]                                       ...(1)
where  c  is a parameter.
As this equation has one arbitrary constant, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
\[\frac{dy}{dx} = 2\left( 2x \right) + c e^{- x^2} ( - 2x)\]
\[ \Rightarrow \frac{dy}{dx} = 4x - 2xc e^{- x^2} . . . \left( 2 \right)\]
From (1) and (2), we get
\[\Rightarrow \frac{dy}{dx} = 4x - 2xy + 4 x^3 - 4x\]
\[ \Rightarrow \frac{dy}{dx} + 2xy = 4 x^3\]
Hence,
\[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\]  is the solution to the differential equation \[\frac{dy}{dx} + 2xy = 4 x^3\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 22: Differential Equations - Exercise 22.02 [Page 17]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 22 Differential Equations
Exercise 22.02 | Q 13 | Page 17

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

\[\frac{d^4 y}{d x^4} = \left\{ c + \left( \frac{dy}{dx} \right)^2 \right\}^{3/2}\]

\[x^2 \left( \frac{d^2 y}{d x^2} \right)^3 + y \left( \frac{dy}{dx} \right)^4 + y^4 = 0\]

Verify that y = \[\frac{a}{x} + b\] is a solution of the differential equation
\[\frac{d^2 y}{d x^2} + \frac{2}{x}\left( \frac{dy}{dx} \right) = 0\]


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


xy (y + 1) dy = (x2 + 1) dx


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

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

Solve the following differential equation:
\[xy\frac{dy}{dx} = 1 + x + y + xy\]

 


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


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

In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (loge 2 = 0.6931).


If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).


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


x2 dy + y (x + y) dx = 0


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

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

y ex/y dx = (xex/y + y) dy


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?


Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

 

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


The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by


Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?


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


If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`


Determine the order and degree of the following differential equations.

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

Solve the following differential equation.

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


Solve the following differential equation.

(x2 − y2 ) dx + 2xy dy = 0


Solve the following differential equation.

y dx + (x - y2 ) dy = 0


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.


Solve:

(x + y) dy = a2 dx


 `dy/dx = log x`


For the differential equation, find the particular solution

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


Find the particular solution of the following differential equation

`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.

Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x

∴ `1/"e"^(2y)  "d"y` = cos x dx

Integrating, we get

`int square  "d"y` = cos x dx

∴ `("e"^(-2y))/(-2)` = sin x + c1

∴ e–2y = – 2sin x – 2c1

∴ `square` = c, where c = – 2c

This is general solution.

When x = `pi/6`, y = 0, we have

`"e"^0 + 2sin  pi/6` = c

∴ c = `square`

∴ particular solution is `square`


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


If `y = log_2 log_2(x)` then `(dy)/(dx)` =


A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×