English

Solve the following differential equation. (x+y)dydx=1

Advertisements
Advertisements

Question

Solve the following differential equation.

`(x + y) dy/dx = 1`

Sum
Advertisements

Solution

`(x + y) dy/dx = 1`

∴ `dy/dx = 1/(x + y)`

∴ `dx/dy = x + y`

∴ `dx/dy − x = y`

∴ `dx/dy + (− 1)x = y`                 ...(I)

The given equation is of the form `dx/dy + Px = Q`

where, P = − 1 and Q = y

∴ `"I.F." = e int ^"pdy" = e int ^("(−1)dy") = e^-"y"`

∴ Solution of the given equation is

`"x (I.F.)" = int "Q (I.F.) dy" + C`

∴ `"xe"^(−"y") = ubrace(int y.e^(−y))_(("I"))   dy + C`        ...(II)

Let I = `int y. e^(−y)  dy`

Using Integration by parts,

I = `y int e^(−y) dy - int [ d/dy y int e^(−y)  dy] dy`

I =  `y (e^(−y))/(-1) − int 1. (e^(−y))/(-1)  dy`

I = `− y. e^(−y) + int e^(−y)  dy`

I = `− y. e^(−y) + e^(−y)/(- 1)  dy`

I = `− y. e^(−y) − e^(−y)`

Putting value of I in (2),

∴ `"xe"^(−"y") = int y.e^(−y)   dy + C`

∴ `"xe"^(−"y") = − y. e^(−y) − e^(−y)  + C`

Dividing by e−y,

∴ x = − y − 1 + Cey

∴  x + y + 1 = Cey

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Differential Equation and Applications - Exercise 8.5 [Page 168]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 8 Differential Equation and Applications
Exercise 8.5 | Q 1.4 | Page 168

RELATED QUESTIONS

Solve the equation for x: `sin^(-1)  5/x + sin^(-1)  12/x = π/2, x ≠ 0`


\[\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \left( c\frac{d^2 y}{d x^2} \right)^{1/3}\]

Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.


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

Differential equation Function
\[x^3 \frac{d^2 y}{d x^2} = 1\]
\[y = ax + b + \frac{1}{2x}\]

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


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


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

Solve the differential equation \[\frac{dy}{dx} = e^{x + y} + x^2 e^y\].

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

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

\[\frac{dy}{dx} = 2 e^x y^3 , y\left( 0 \right) = \frac{1}{2}\]

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

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

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.


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

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 \[x = \frac{\pi}{2}\]


Find the equation of the curve which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\]  at any point (x, y) on it.


The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).


The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by


Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]


Which of the following differential equations has y = C1 ex + C2 ex as the general solution?


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


Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.


For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1


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 differential equation:

dr = a r dθ − θ dr


Solve the differential equation xdx + 2ydy = 0


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


Solve the differential equation

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


Solve the differential equation

`x + y dy/dx` = x2 + y2


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×