Differential Equations

e? dy = e? /α dx
⇒−e? =e? /α+c
y (ln2)=ln2 and y (0)=−ln2
⇒−2=−1/α+c
⇒c=−2−1/α
⇒e? = 1/α e? −2−1/α
⇒−e? ² = 1/α e? ²−2−1/α
⇒2? ¹=3/α
⇒α=2

dy/dx + (2? (2? -1)/ (2? ¹ (2? -1) = 0
x, y > 0, y (1) = 1
dy/dx = – (2? (2? -1)/ (2? (2? -1)
∫ (2? -1)/2? dy = –∫ (2? -1)/2? dx
log? (2? -1)/log?2 = – log? (2? -1)/log?2 + log? c/log?2
Taking log of base 2.
∴ y = 2 – log?3

I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4

Given kx^ (k-1) + k * y^ (k-1) * dy/dx = 0.
dy/dx = - (kx^ (k-1) / (ky^ (k-1) = - (x/y)^ (k-1).
The provided solution has dy/dx + (x/y)^ (k-1) = 0.
It seems to relate to k-1 = -1/3, which implies k = 1 - 1/3 = 2/3.

dy/dx + 2y tan (x) = sin (x)
I.F. = e^ (∫2tan (x)dx) = e^ (2ln (sec (x) = sec² (x)
Solution is y sec² (x) = ∫sin (x)sec² (x)dx = ∫sec (x)tan (x)dx
⇒ y sec² (x) = sec (x) + c
y (π/3) = 0 ⇒ 0 * sec² (π/3) = sec (π/3) + c ⇒ 0 = 2 + c ⇒ c = -2.
∴ y = (sec (x) - 2) / sec² (x)
Now let g (t) = (t - 2)/t² = 1/t - 2/t² for |t| ≥ 1.
g' (t) = -1/t² + 4/t³
g' (t) = 0 ⇒ t = 4.
g' (t) = 2/t³ - 12/t? g' (4) < 0, hence maximum.
∴ g (t)max = g (4) = (4 - 2)/4² = 2/16 = 1/8.

dy/dx + (tanx)y = sinx. This is a linear differential equation.
Integrating Factor (I.F.) = e^ (∫tanx dx) = e^ (ln|secx|) = secx.
The solution is y * I.F. = ∫ (sinx * I.F.) dx + C.
y * secx = ∫ (sinx * secx) dx = ∫tanx dx = ln|secx| + C.
Given y (0) = 0.
0 * sec (0) = ln|sec (0)| + C => 0 = ln (1) + C => C = 0.
So, y * secx = ln (secx).
y = cosx * ln (secx).
At x = π/4:
y = cos (π/4) * ln (sec (π/4) = (1/√2) * ln (√2) = (1/√2) * (1/2)ln (2) = ln (2) / (2√2).