Differential Equations Part 1

Given, y

1

4

+y−14=2x
⇒(y

1

4

+y−

1

4

)2=(2x)2
⇒(y

1

4

+y−

1

4

)2=4x2
Differentiating w.r.t. x, we get

1

4y

(y

1

4

−y−

1

4

)

dy

dx

=2
⇒(y

1

4

−y−

1

4

)

dy

dx

=8y ...(i)
Now, y

1

4

−y−

1

4

=√(y

1

4

+y−

1

4

)2−4
⇒y

1

4

−y−

1

4

=2√x2−1 ...(ii)
⇒√x2−1)‌

dy

dx

=4y [using Eqs. (i) and (ii)]
Squaring on both sides,
(x2−1)(

dy

dx

)2=16y2
Again, differentiating w.r.t. x
(x2−1).2

dy

dx

.

d2y

dx2

+2x(

dy

dx

)2=32y

dy

dx

On dividing by

2dy

dx

, we get
(x2−1)

d2y

dx2

+x

dy

dx

=16y
or(x2−1)

d2y

dx2

+x

dy

dx

−16y=0
Comparing with
(x2−1)

d2y

dx2

+αx

dy

dx

+βy=0
α =1, β = − 16
∴|α−β|=|1+16|=17