Question

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

x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Answer 1

Given differential equation,

x²dy + (xy + y²) dx = 0

or `dy/dx = (- (xy + y^2))/x^2 = - (y/x) - (y/x)^2`

Clearly, this is a homogeneous differential equation.

∴ Putting y = vx 

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

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

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

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

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

On integrating

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

`=> 1/2 [log v - log (v + 2)] = - log x + log C`

`=> log (v/(v + 2)) = 2 log (C/x)`

`=> log (v/(v + 2)) = log (C/x)^2`

`=> v/(v + 2) = (C/x)^2`

On substituting `y/x` in place of v

`(y/x)/(y/x + 2) = C^2/x^2`

x²y = C²(y + 2x)     ...(ii)

Given, Putting x = 1, y = 1 in equation (ii),

1 = C² (1 + 1) ⇒ C² = `1/3`

Putting this value of C² in equation (ii),

`x^2y = 1/3(y + 2x)`

y + 2x = 3x²y