Question

(x2+ y² + 1) dx - 2xy dy = 0
Solve differential equation​

Answer 1

Step-by-step explanation:

How do I solve (x²+y²) dx + 2xy dy= 0?

(x2+y2)dx+2xydy=0⇒dydx=−x2+y22xy

This is homogeneous equation

Let,y=vx⇒dydx=v+xdvdx⇒v+xdvdx=−x2+v22x.vx⇒xdvdx=−v−1+v22v⇒xdvdx=−1+3v22v⇒dxx=−2v1+3v2⇒∫dxx=−∫2v1+3v2⇒logx=−13log(1+3v2)+logc⇒logx+13log(1+3v2)=logc⇒logx(1+3v2)13=logc⇒x(1+3v2)13=c⇒x(1+3y2x2)13=c⇒x(x2+3y2x2)13=c

Pulmonary arterial hypertension and telemedicine: What to know.

Use a substitution:⎧⎩⎨⎪⎪u=yxy=uxdy=udx+xdu

(x2+y2)dx+2xydy=0

(x2+(ux)2)dx+2x(ux)(udx+xdu)=0

(x2+u2x2+2u2x2)dx+2ux3du

x2(1+3u2)dx=−2ux3du

1xdx=−2u1+3u2du

∫1xdx=∫−2u1+3u2du

ln|x|=−13ln|1+3u2|+C0

3ln|x|=−ln|1+3u2|+C1

ln|x3|+ln∣∣1+3⋅y2x2∣∣=C1

ln|x3+3xy2|=C1

x3+3xy2=C

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