For the differential equation find a particular solution satisfying the given condition: (x + y) dy + (x – y) dx = 0; y = 1 when x = 1 - Mathematics

Question

For the differential equation find a particular solution satisfying the given condition:

(x + y) dy + (x – y) dx = 0; y = 1 when x = 1

Sum

Solution

given (x + y) dy + (x – y) dx = 0

`=> dy/dx = (y - x)/(y + x)` ...(i)

∵ The powers of the numerator and denominator are the same so this is a homogeneous differential equation.

∴ Putting y = vx

`dy/dx = v + x (dv)/dx` ...(in equation (i))

`=> v + x (dv)/dx = (vx - x)/(vx + x)`

`=> x (dv)/dx = (v - 1)/(v + 1) - v`

`x (dv)/dx = (v - 1 - v^2 - 1)/(v + 1)`

`= - (v^2 + 1)/(v + 1)`

`(v + 1)/(v^2 + 1)dv = - 1/x dx`

On integrating,

`=> 1/2 int (2v)/(v^2 + 1)dv + int 1/(v^2 + 1) dv = - int 1/x dx`

`1/2 log (v^2 + 1) + tan^-1 (v) = - log x + C`

log (v² + 1) + 2 tan^-1 (v) = - 2 log x + 2C

So on putting `y/x` in place of v,

`log ((y^2 + x^2)/x^2) + 2 tan^-1 (y/x) = - log x^2 + 2C`

`log (x^2 + y^2) - log x^2 + 2 tan^-1 (y/x) = - log x^2 x + 2C`

`log (x^2 + y^2) + 2 tan^-1 (y/x) = 2C` ....(ii)

Given y = 1 and x = 1

log (1² + 1²) + 2 tan^-1 (1) = 2C

log 2 + 2 tan^-1 (1) = 2C

2C = log 2 + 2 `xx pi/4 = log 2 + pi/2`

Putting this value of C in equation (ii),

`log (x^2 + y^2) + 2 tan^-1 (y/x) = pi/2 + log 2`

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Methods of Solving First Order, First Degree Differential Equations - Homogeneous Differential Equations

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Chapter 9: Differential Equations - Exercise 9.5 [Page 406]

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NCERT Mathematics Part 1 and 2 [English] Class 12

Chapter 9 Differential Equations
Exercise 9.5 | Q 11 | Page 406

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