Differential Equations

x³dy + xydx = 2ydx + x²dy
=> (x³-x²)dy = (2-x)ydx
=> dy/y = (2-x)/ (x² (x-1) dx
=> ∫ (dy/y) = ∫ (2-x)/ (x² (x-1)dx
Let (2-x)/ (x² (x-1) = A/x + B/x² + C/ (x-1)
=> 2-x = A (x-1) + B (x-1) + Cx²
=> C=1, B=-2 and A=-1
=> ∫ (dy/y) = ∫ (-1/x - 2/x² + 1/ (x-1)dx
=> lny = -lnx + 2/x + ln|x-1| + C
∴ y (2) = e
=> 1 = -ln2 + 1 + 0 + C
=> C = ln2
=> lny = -lnx + 2/x + ln|x-1| + ln2
at x = 4
=> lny (4) = -ln4 + 1/2 + ln3 + ln2
=> lny (4) = ln (3/4) + 1/2 = ln (3/2)e^ (1/2)
=> y (4) = (3/2)e^ (1/2)

(1 + e? ) (1 + y²) dy/dx = y²
⇒ (1 + y? ²)dy = ( e? / (1 + e? ) ) dx
⇒ (y - 1/y) = ln (1 + e? ) + c
∴ It passes through (0,1) ⇒ c = -ln2
⇒ y² = 1 + yln ( (1+e? )/2 )

Kindly consider the following Image

(2+sin x)/ (y+1) dy/dx = -cosx, y>0
⇒ dy/ (y+1) = -cosx/ (2+sinx) dx
By integrating both sides:
ln|y+1| = -ln|2+sinx|+lnK
⇒ y+1 = K/ (2+sinx) (y+1>0)
⇒ y (x) = K/ (2+sinx) - 1
Given y (0)=1 ⇒ 1=K/2-1 ⇒ K=4
So, y (x)=4/ (2+sinx)-1
a=y (π)=1
b=dy/dx|x=π = -cosx/ (y (x)+1)|x=π = 1
So, (a, b)= (1,1)