Solution
Step 1: Equation ko simplify karein
Di gayi equation:
(3y3−5x2y)dx+(2x3−2xy2)dy=0
Kyunki har term ki degree 3 hai, yeh ek homogeneous differential equation hai.
Step 2: Substitution y=vx rakhein
Differentiate karne par: dxdy=v+xdxdv
Equation mein values rakhne par:
v+xdxdv=2x3−2x(vx)25x2(vx)−3(vx)3
xdxdv=2−2v25v−3v3−2v+2v3=2−2v23v−v3
Step 3: Variables separate karke Integrate karein
LHS ko solve karne ke liye, dhyaan dein ki d(3v−v3)=(3−3v2)dv.
Hum LHS ko likh sakte hain: 32∫3v−v33−3v2dv
v=y/x wapas rakhne par:
(3xy−x3y3)2=C′x3⟹x6(3yx2−y3)2=C′x3
Step 4: Boundary condition y(1)=1 apply karein
Humein equation mili: (3yx2−y3)2=4x9
Dono taraf square root lene par: 3yx2−y3=±2x9/2
Step 5: y(2) ki value find karein
x=2 rakhne par:
3y(2)(4)−(y(2))3=±2(2)9/2
Humein ∣(y(2))3−12y(2)∣ ki value nikaalni hai:
∣−(12y(2)−(y(2))3)∣=∣±322∣=322
Correct Option: (2) 322