English

( X 2 + 1 ) D Y D X = 1

Advertisements
Advertisements

Question

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

Solution

We have, 
\[\left( x^2 + 1 \right)\frac{dy}{dx} = 1\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x^2 + 1}\]
\[ \Rightarrow dy = \left( \frac{1}{x^2 + 1} \right)dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\left( \frac{1}{x^2 + 1} \right)dx\]
\[ \Rightarrow y = \tan^{- 1} x + C\]
\[\text{ So,} y = \tan^{- 1} x + \text{ C is defined for all }x \in R . \]
\[\text{ Hence, }y = \tan^{- 1} x + \text{ C, where }x \in R, \text{ is the solution to the given differential equation }.\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 21: Differential Equations - Exercise 22.05 [Page 34]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.05 | Q 4 | Page 34

RELATED QUESTIONS

\[\sqrt[3]{\frac{d^2 y}{d x^2}} = \sqrt{\frac{dy}{dx}}\]

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

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


Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]

 


\[\frac{dy}{dx} + 2x = e^{3x}\]

\[\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}\]

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

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

C' (x) = 2 + 0.15 x ; C(0) = 100


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

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

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

 


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

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

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

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

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


Solve the following initial value problem:-

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


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 decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.


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.


The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.


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


The solution of the differential equation y1 y3 = y22 is


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


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is


The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]


Form the differential equation representing the family of curves y = a sin (x + b), where ab are arbitrary constant.


Solve the differential equation:

`"x"("dy")/("dx")+"y"=3"x"^2-2`


Solve the following differential equation.

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


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


Solve the following differential equation.

dr + (2r)dθ= 8dθ


Choose the correct alternative.

Bacteria increases at the rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in


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

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c


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`


Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.


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


The value of `dy/dx` if y = |x – 1| + |x – 4| at x = 3 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×