Differential Equations Part 1

Given, (2xy2−y)dx+xdy=0
⇒‌‌‌

dy

dx

−‌

y

x

=−2y2
⇒‌‌‌

−1

y2

‌

dy

dx

+‌

1

xy

=2 ...(i) [divide by y2 ]
Let ‌

1

y

=v, then −‌

1

y2

⋅‌

dy

dx

=‌

dv

dx

, putting in Eq. (i)
‌

dv

dx

+v(‌

1

x

)=2 (this is a linear form)
Now, integrating factor (IF) =e∫‌

1

x

dx=elog‌x=x
∴‌‌(IF)v=∫2.(IF)dx=∫2xdx=2‌

x2

2

+C
∴‌‌(IA)v=x2+C
Put v=‌

1

y

, this gives
x2+c=‌

x

y

Now, first find point of intersection of lines
2x−3y=1 and 3x=−2y+8 by elimination method, we get x=2, y=1
∴ The curve x2+c=‌

1

y

passes through (2,1).
Put x=2,y=1, we get c=−2
‌

x

y

=x2−2
or y=‌

x

x2−2

Put x=1, we get y(1)=‌

1

1−2

=−1
∴ |y(1)|=1