Step 1: k ki value nikalna
Maan lijiye point P=(x,y). Tangent ki equation hoti hai:
Section formula ke anusar, agar P line AB ko 1:k mein divide karta hai:
x=1+k1(0)+k(XA)⟹x=1+kk(x−dy/dxy)
Ise simplify karne par humein curve ki basic equation milti hai: xky=c.
Humein do points diye gaye hain: (1,1) aur (101,100).
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(1)k(1)=c⟹c=1
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(101)k(100)=1⟹10−k⋅102=100
Step 2: Differential Equation ko solve karna
Ab k=2 ko equation mein rakhte hain:
edxdy=2x+22⟹edxdy=2x+1
Dono taraf loge lene par:
Ab integrate karte hain:
Integration by parts (∫ln(u)du=ulnu−u) ka use karke:
y=2(2x+1)ln(2x+1)−(2x+1)+C
Condition y(0)=k=2 ka use karke C nikalte hain:
2=2(1)ln(1)−1+C⟹2=−21+C⟹C=25
Toh final equation hui:
y(x)=2(2x+1)ln(2x+1)−(2x+1)+5
Step 3: Final Answer nikalna
Humein 4y(1)−5ln3 ki value chahiye. Pehle y(1) nikalte hain:
y(1)=2(2(1)+1)ln(2(1)+1)−(2(1)+1)+5
Ab expression mein value rakhte hain:
