Differential Equations Part 1

Given, 2(x2+x5/4)dy−y⋅(x+x1/4)dx=
2x9/4dx, where x>0
After rearranging, we get
‌

dy

dx

−‌

y

2x

=‌

x9/4

x5/4(x3/4+1)

This is of the form ‌

dy

dx

+Py=Q, where P and Q are constants or function of x.
∴ Integrating factor (IF) =e∫Pdx
=e∫−‌

1

2x

dx=e‌

−1

2

‌∫xdx
=e−‌

1

2

‌log‌x=elog(x)−1/2=‌

1

x1/2

Its solution is
y×(IF)=∫Q×(IF)dx
⇒‌‌y×‌

1

x1/2

=∫‌

x1/2dx

(x3/4+1)

...(i)
Putting x=z4
⇒‌‌dx=4z3⋅dz
RHS of Eq. (i) becomes,
∫‌

z2⋅4z3

(z3+1)

⋅dz=4‌∫‌

z2(z3+1−1)

(z3+1)

dz
=4[∫‌

z2(z3+1)

(z3+1)

dz−∫‌

z2

(z3+1)

dz]
=4[‌

z3

3

−‌

1

3

⋅∫‌

3z2

z3+1

dz]
=4[‌

z3

3

−‌

1

3

⋅log‌|z3+1|]
(∵∫‌

f′(x)

f(x)

dx=log‌|f(x)|+C)
=‌

4z3

3

−‌

4

3

‌log‌|z3+1|+C
∴ Eq. (i) becomes,
Since, this passes through (1,1−‌

4

3

loge2)
Then, ⇒C=1−‌

4

3

⇒C=‌

−1

3

Hence, ∵x>0
Now, putting x=16, we get
y(16)=‌

4

3

×32−‌

4

3

×4‌log‌9−‌

4

3

=‌

124

3

−‌

32

3

‌log‌3
=4(‌

31

3

−‌

8

3

‌log‌3)