**Solve the following differential equation.**

`dy /dx +(x-2 y)/ (2x- y)= 0`

#### Solution

`dy /dx +(x-2 y)/ (2x- y)= 0` ....(i)

Put y = tx ...(ii)

Differentiating w.r.t. x, we get

`dy/dx = t + x dt/dx `...(iii)

Substituting (ii) and (iii) in (i), we get

`t + x dt/dx + (x-2tx)/(2x-tx) = 0`

∴`x dt/dx +t + (1-2t)/2-t = 0`

∴`x dt/dx + (2t - t^2+1-2t)/2-t = 0`

∴`x dt/dx + (1-t^2)/(2-t )= 0`

∴ `x dt/dx = - (1-t^2)/(2-t )`

∴ = `(2-t)/(1-t^2)dt = dx/x`

∴ `(2-t)/(t^2-1)dt = dx/x`

Integrating on both sides, we get

`int (2-t)/(t^2-1) dt = int dx/x`

∴ `int (2-t)/((t+1)(t-1)) dt = int dx/x`

Let `2-t/((t+1)(t-1)) = A/(t+1)+ B/(t-1)`

∴ 2 - t = A(t -1) + B(t + 1)

Putting t = 1, we get

∴ 2 -1 = A(1 -1) + B(1 + 1)

∴ B = `1 /2`

Putting t = -1, we get

2 -(-1) = A(-1 -1) + B(-1 + 1)

∴ A = `(-3)/2`

∴ `int(-3/2)/(t+1)dt +int(1/2)/(t-1) dt = intdx/x`

∴`(-3)/2 int 1/(t+1)dt + 1/2int 1/(t-1) dt = int dx/x`

∴`(-3)/2 log|t+1| + 1/2 log |t-1| = log |x| + log |c_1|`

∴ `-3 log |(y+x)/x| + log|(y-x)/x| = 2log |x| + 2 log |c_1|`

∴ -3 log |y+x| + 3 log |x| + log | y -x| - log |x|

= 2 log |x| + 2 log |c_{1}|

∴ log |y - x| = 3 log |y+x|+ 2 log |c_{1}|

∴ log |y- x |= log |( y+ x )^{3}|+ log | c_{1}^{2}|

∴ log | y - x| = log | c_{1}^{2} ( x+y)^{3}|

∴ (y - x) = c(x + y)^{ 3} … |c_{1}^{2} c|

#### Notes

Answer given in the textbook is `log |(x+y)/(x-y)| - 1/2 log | x^2 - y^2| + 2 log x = log c.`

However, as per our calculation it is ‘(y -x) = c(x+y)^{3}.