z = 1/y² → y² = 1/z
Differentiate w.r.t x , dz/dx = -2/y³ dy/dx
dy/dx = -y³/2 dz/dx

Given : dy/dx = y tan x + y³ tan³ x
Sub dy/dx = -y³/2 dz/dx,

-y³/2 dz/dx = y tan x + y³ tan³x

Multiply both sides by y,

-y⁴/2 dz/dx = y² tan x + y⁴ tan³x

Sub y² = 1/z,

-1/2z² dz/dx = 1/z tan x + 1/z² tan³ x

Multiply both sides by -2z²,

dz/dx = -2z tan x - 2 tan³ x
dz/dx + 2z tan x = -2tan³ x (shown)

dz/dx + 2z tan x = -2tan³ x

Integrating factor = e^(∫2tan x dx)

= e^(∫ 2sin x / cos x dx)

= e^(-2ln|cos x|)

= e^(ln|cos x|-²)

= |cos x|-²

= 1 / |cos x|²

= 1 / cos²x (since |cos x|² ≥ 0 so we can write it as cos²x, which is also ≥ 0 and we can drop the modulus sign.


Multiply the equation above by 1/cos²x ,

1/cos²x dz/dx + 2z/cos²x tan x = -2/cos²x tan³ x

sec²x dz/dx + (2sec²x tanx)z = -2sec x tan³x



Integrate both sides with respect to x,

∫ (sec²x dz/dx + (2sec²x tanx)z ) dx = ∫-2sec x tan³x dx

sec²x z = -½tan⁴x + A, A is a constant


Sub z = 1/y² ,

1/y² sec²x = -½tan⁴x + A

1/y² (1/cos²x) = A - ½tan²x (sin²x/cos²x)

1/y² = Acos²x - ½tan²x sin²x

2/y² = 2Acos²x - tan²x sin²x

y² = 2 / (2Acos²x - tan²x sin²x)

y² = 2 / (Bcos2x - tan²x sin²x) , where B = 2a


(You can choose to find y by ± square rooting the RHS if you want, but leaving it in this form is usually alright.)

Q2

2x(4x² + 1) dy/dx - (4x² + 1) y = 2x^(5/2) dy/dx - y/2x = x³/² / (4x² + 1)


Integrating factor :

e^(∫ -1/2x dx)
= e^(-½ ln|x|)
= e^(ln|x|-¹/²)
=|x|-¹/²
= 1/√|x|
= 1/√x

(RHS of the equation = 2x^(5/2) = 2(√x)^5 = 2√(x^5) . In order for real values of this to exist, x ≥ 0 so the modulus sign caj be dropped)

Then, multiply the equation by 1/√x

1/√x dy/dx - y/(2x√x) = x³/² / (√x(4x² + 1))

x-¹/² dy/dx - ½y x-³/² = x/(4x² + 1)

Integrate both sides with respect to x,

∫ (x-¹/² dy/dx - ½y x-³/²) dx = ∫ x/(4x² + 1) dx

x-¹/² y = ⅛ln(4x² + 1) + C , C is a constant

(No need for modulus since 4x² + 1 > 0 for all real x)

y = ⅛x¹/² ln(4x² + 1) + Cx¹/²

y = ⅛√x ln(4x² + 1) + C√x

hi how did u know to integrate the factor e^ (...) ?

Because these two questions are not variable-separable. We are not able to bring the y to one side and x to the other side.

So we have to use the integrating factor to make the left hand side in the form which can be directly integrated back. (I.e reverse of the product rule)


Have you been taught the integrating factor?

https://tutorial.math.lamar.edu/Classes/DE/Linear.aspx