Differential Equations Part 2

Given,
x‌

dy

dx

−y=√y2+16x
⇒x‌

dy

dx

=y+√y2+16x
⇒‌

dy

dx

=‌

y

x

+√(‌

y

x

)2+16
This is a homogenous different equation.
Let ‌

y

x

=v
⇒y=vx
⇒‌

dy

dx

=v+x‌

dv

dx

∴v+x‌

dv

dx

=v+√v2+16
⇒x‌

dv

dx

=√v2+16
⇒‌

dv

√v2+16

=‌

dx

x

Integrating both sides, we get
∫‌

dv

√v2+16

=∫‌

dx

x

⇒ln|v+√v2+16

‌

‌

|=ln‌x+ln‌c
⇒v+√v2+16=cx
Now putting, v=‌

y

x

, we get
‌

y

x

+√‌

y2

x2

+16=cx
⇒‌

y

x

+√‌

y2+16x2

x2

=cx
⇒y+√y2+16x2=cx2......
Given, y(1)=3
∴ When x=1 then y=3.
Putting in equation (1) we get,
3+√9+16=c.1
⇒c=8
∴ Solution of equation,
y+√y2+16x2=8x2
Now, y(2) means when x=2 then y= ?

∴y+√y2+16×4=8×4
⇒y=15