Question

The solution of the differential equation, `(dy)/(dx)` = (x – y)², when y (1) = 1, is ______.

Options

• `log_e |(2 - x)/(2 - y)|` = x – y

• `-log_e |(1 - x + y)/(1 + x - y)|` = 2(x – 1)

• `-log_e |(1 + x - y)/(1 - x + y)|` = x + y – 2

• `log_e |(2 - y)/(2 - x)|` = 2(y – 1)

Answer 1

The solution of the differential equation, `(dy)/(dx)` = (x – y)², when y (1) = 1, is `underlinebb(-log_e |(1 - x + y)/(1 + x - y)| = 2(x - 1))`.

Explanation:

The given differential equation

`(dy)/(dx)` = (x – y)²    ...(i)

Let x – y = t 

`\implies 1 - (dy)/(dx) = (dt)/(dx)` 

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

Now, from equation (i)

`(1 - (dt)/(dx))` = (t)²

`\implies` 1 – t²  = `(dt)/(dx)`

 `\implies intdx = int (dt)/(1 - t^2)`

`\implies` x = `1/(2 xx 1) log|(1 + t)/(1 - t)| + c`

`\implies` x = `1/2 log|(1 + x - y)/(1 - (x - y))| + c`  ...(ii)

`\implies` x = `1/2 log|(1 + x - y)/(1 - (x - y))| + c`

 From given condition y(1) = 1

1 = `1/2 log|(1 + 1 - 1)/(1 - (1 - 1))| + c`

`\implies` c = 1   ...(ii)

∴ From equation

x = `1/2 log|(1 + x - y)/(1 - x + y)| + 1`

`\implies` 2(x – 1) = `-log|(1 - x + y)/(1 + x - y)|`