Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

Question

asked Oct 27, 2020 58.5k views

1 Answer

Iskar Jarak · Answer 1 · 2020-10-31T06:16:17+0000

Answer:

(y)/(x)-2ln((y)/(x)+1)=lnx+C

Explanation:

Given differential equation,

(x^2 + y^2) dx + (x^2 - xy) dy = 0

$\implies (dy)/(dx)=-(x^2 + y^2)/(x^2 - xy)----(1)$

Let y = vx

Differentiating with respect to x,

(dy)/(dx)=v+x(dv)/(dx)

From equation (1),

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

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

v+x(dv)/(dx)=-(1 + v^2)/(1 - v)

v+x(dv)/(dx)=(1 + v^2)/(v-1)

x(dv)/(dx)=(1 + v^2)/(v-1)-v

x(dv)/(dx)=(1 + v^2-v^2+v)/(v-1)

x(dv)/(dx)=(v+1)/(v-1)

(v-1)/(v+1)dv=(1)/(x)dx

Integrating both sides,

$\int{(v-1)/(v+1)}dv=\int{(1)/(x)}dx$

$\int{(v-1+1-1)/(v+1)}dv=lnx + C$

$\int{1-(2)/(v+1)}dv=lnx + C$

v-2ln(v+1)=lnx+C

Now, y = vx ⇒ v = y/x

$\implies (y)/(x)-2ln((y)/(x)+1)=lnx+C$