Question

Solve the differential equation: `y - x dy/dx = 0`.

Solution: given equation is `y - x dy/dx = 0`

Separating the variables, we get

`dx/square = dy/square`

Integrating, we get

`int dx/square = int dy/square + c`

∴ log x = `square + c`

∴ log x − log y = log c₁, where c = log c₁

∴ `log (x/y) = log c_1`

∴ `x/square = c_1`

Hence, the required solution is x = c₁y.

Answer 1

Given equation is `y - x dy/dx = 0`

Separating the variables, we get

\[\frac{dx}{\boxed{x}}\] = \[\frac{dy}{\boxed{y}}\]

Integrating, we get

\[\int{\frac{dx}{\boxed{x}}}\] = \[\int{\frac{dy}{\boxed{y}}}\] + c

∴ log x = \[\boxed{\text{log y}}\] + c

∴ log x − log y = log c₁, where c = log c₁

∴ `log (x/y) = log c_1`

∴ \[\frac{x}{\boxed{y}}\] = c₁

Hence, the required solution is x = c₁y.