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NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice

The solution of the differential equation 1 x 2 d y d x 2 + 6 = 5 x d y d x is y = λ x 2 + c (where, c is an arbitrary constant). The sum of all the possible value of λ is

Options

A3 2
B5 2
C2 5
D2

Correct answer

B. 5 2

Step-by-step solution

Given equation is d y d x 2 - 5 x d y d x + 6 x 2 = 0 or d y d x - 3 x d y d x - 2 x = 0 ⇒ y = 3 2 x 2 + k or ⇒ y = x 2 + μ

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