A function y = f (x) satisfies (x + 1) . f ′ (x) – 2 (x 2 + x) f (x) = e ^ x ^ 2 ( x +1)…
Mathematics · JEE Advanced · NTA Exams — Differential Equations
A function y = f (x) satisfies
(x + 1) . f ′ (x) – 2 (x2 + x) f (x) = \(\frac{\mathrm{e}^{\mathrm{x}^{2}}}{(\mathrm{x}+1)}\) , \(\forall x>-1\)
If f (0) = 5 , then f (x) is
(x + 1) . f ′ (x) – 2 (x2 + x) f (x) = \(\frac{\mathrm{e}^{\mathrm{x}^{2}}}{(\mathrm{x}+1)}\) , \(\forall x>-1\)
If f (0) = 5 , then f (x) is
- \(\left(\frac{3 x+5}{x+1}\right) \cdot e^{x^{2}}\)
- \(\left(\frac{6 x+5}{x+1}\right) \cdot e^{x^{2}}\)
- \(\left(\frac{6 x+5}{(x+1)^{2}}\right) \cdot e^{x^{2}}\)
- \(\left(\frac{5-6 x}{x+1}\right) \cdot e^{x^{2}}\)
Answer
(B) ( 6 x+5 x+1 ) e^ x^ 2
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