Question

The particular solution of the differential equation (2x – 2y + 3) dx – (x – y + 1) dy = when x = 0, y = 1 is ______.

पर्याय

• 2x – y – log (x – y + 2) + 1 = 0

• x – 2y – log (x – y + 2) + 2 = 0

• 2x + y – log (x – y + 2) – 1 = 0

• x – y – log (x – y + 2) + 1 = 0

Answer 1

The particular solution of the differential equation (2x – 2y + 3) dx – (x – y + 1) dy = when x = 0, y = 1 is 2x – y – log (x – y + 2) + 1 = 0.

Explanation:

We have, (2x – 2y + 3) dx – (x – y + 1) dy = 0

`\implies dy/dx = (2(x - y) + 3)/(x - y + 1)`

Put x – y = v

`\implies 1 - dy/dx = (dv)/dx`

`\implies dy/dx = 1 - (dv)/dx`

`\implies 1 - (dv)/dx = (2v + 3)/(v + 1)`

`\implies (dv)/dx = 1 - (2v + 3)/(v + 1) = (-(v + 2))/(v + 1)`

`\implies ((v + 1)/(v + 2))dv` = – dx

`\implies (1 - 1/(v + 2))dv` = – dx  ...(i)

On integrating both sides of equation (i), we get

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

`\implies` v – log (v + 2) = – x + C

`\implies` x – y – log (x – y + 2) = – x + C  ...(ii)

On putting x = 0, y = 1 in equation (ii), we get C = – 1

∴ x – y – log (x – y + 2) = – x – 1

`\implies` 2x – y – log (x – y + 2) + 1 = 0