JEE Mains · Maths · STD 12 - 9. differential equations
If \(x^{3} d y+x y d x=x^{2} d y+2 y d x ; y(2)=e\) and \(x\) \(>1,\) then \(y (4)\) is equal to
- A \(\frac{3}{2}+\sqrt{ e }\)
- B \(\frac{3}{2} \sqrt{ e }\)
- C \(\frac{1}{2}+\sqrt{ e }\)
- D \(\frac{\sqrt{ e }}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{2} \sqrt{ e }\)
Step-by-step Solution
Detailed explanation
\(x ^{3} dy + xy dx = x ^{2} dy +2 y dx\) \(\Rightarrow \quad dy \left( x ^{3}- x ^{2}\right)= dx (2 y - xy )\) \(\Rightarrow \quad-\int \frac{1}{ y } dy =\int \frac{ x -2}{ x ^{2}( x -1)} dx\) \(\Rightarrow\) Where \(A =1, B =+2, C =-1\)…
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