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)

f" (0)=10, f" (0+)=2λ. λ=5.

e? (dy/dx) - 2e? sinx + sinxcos²x = 0
d/dx (e? ) - (2sinx)e? = -sinxcos²x
I.F. = e^ (-∫2sinxdx) = e²cosx
Solution: e? e²cosx = -∫e²cosx sinx cos²x dx
Let cosx=t, -sinxdx=dt
∫e²? t²dt = e²? t²/2 - ∫2te²? /2 dt = e²? t²/2 - [te²? /2 - ∫e²? /2 dt] = e²? t²/2 - te²? /2 + e²? /4 + C
e^ (y+2cosx) = e²cosxcos²x/2 - e²cosxcosx/2 + e²cosx/4 + C
y (π/2)=0 ⇒ e? = 0 + 0 + e? /4 + C ⇒ C=3/4
y = ln (e²cosx (cos²x/2-cosx/2+1/4)+3/4e? ²cosx)
y (0) = ln (e² (1/2-1/2+1/4)+3/4e? ²) = ln (e²/4+3/4e? ²)
This seems very complex. The solution provided leads to α=1/4, β=3/4. 4 (α+β)=4.

√ (1+x²) (1+y²) + xy (dy/dx)=0.
√ (1+x²)/x dx + √ (1+y²)/y dy = 0.
√ (1+x²)+½ln| (√ (1+x²)-1)/ (√ (1+x²)+1)|+√ (1+y²)=C.