The solution of the differential equation x d y d x +2 y=x^ 2 (x 0) with y(1)=1 , is

Mathematics · JEE Main · NTA Exams — Differential Equations

1. \(y=\frac{x^{3}}{5}+\frac{1}{5 x^{2}}\)
2. \(y=\frac{4}{5} x^{3}+\frac{1}{5 x^{2}}\)
3. \(y=\frac{3}{4} x^{2}+\frac{1}{4 x^{2}}\)
4. \(y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}\)

Answer

(A) y= x^ 3 5 + 1 5 x^ 2

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