The solution of the differential equation dy dx y empty prime x minus y2 empty x ...........

Mathematics (NDA) Differential Equations

The solution of the differential equation

dy / dx = y∅ ′ (x)−y² / ∅(x) is -

y = x / ∅(x)+c

y = ∅(x) / x + c

. y = ∅(x)+c / x

y = ∅(x) / x+c

dy / dx = y ∅′ (x)− y² / (x) = y (∅′(x) / ∅(x) )– y² / ∅(x)

Dividing by y² on both sides -

1/y² dy / dx = 1/y ( ∅′(x) ∅(x) ) - 1 / ∅(x)

or 1/y² dy/dx – 1/y( ∅′(x)/∅(x) ) = - 1 / ∅(x)

Let −1/y =z

1/y² dy/dx = dz/dx

Above equation becomes:

dz/dx - (∅′ (x) / ∅(x) ) z = - 1 / ∅(x)

I.F = _e∫ ∅′(x) / ∅(x) dx = _e^log∅(x) = ∅(x)

Solution of above equation is:

z. ∅’(x) = ∫ −1 / ∅(x) ×∅(x) dx

−1 /y ∅(x) = -x

or y = ∅(x) /x + c

Submit Error