solve this differntial equation?

Question

y'=2y / x

y(1)=2

Answer 1

dy/dx=2y/x
dy/y=2.dx/x
ln(y)=2ln(x)+ln(c)
y=c.x^2

y(1)=2
2=c.1
c=2

y=2x^2

Answer 2

i'm assuming that your first equation has an blunders in it and your 2nd one is authentic. it is, the final equation is (x^2 + 3xy +y^2)dx - x^2 dy = 0 is the final assertion. it is homogeneous. enable y = v x, so dy = x dv + v dx. replace for v, removing y and dy everywhere. Then simplify your equation. Then it is going to likely be accessible to split variables. Separate them, combine and remedy for v and finally for y.

Answer 3

dy/dx=2y/x
dy/y=2dx/x
Integrating both sides
lnlyl=2lnlxl+lnc
No harm made if I put
instead c=lnc
lnlyl=ln(cx^2)
y=cx^2
I ignore abs.value of y
because c can take
both possitive or negative
values,
For x=1-->y=2
-->c=2
Then y=2x^2

Answer 4

dy/dx=2y/x
dy/y=2.dx/x
ln(y)=2ln(x)+ln(c)
y=c.x^2
but given, y(1)=2
2=c.1
c=2

y=2x^2

Answer 5

dy/dx = 2y / x
(1/2).∫ (1/y).dy = ∫ (1/x).dx
(1/2).log y = log x + C
(1/2).(log 2) = log1 + C
C = (1/2) log 2
(1/2) log y = log x + (1/2) log 2
log y = 2 log x + log 2
log y = log x² + log 2
y = 2x²