JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

y = cx - {c^2}

, is the general solution of the differential equation

View answer and explanation

Correct option: A

{\left( {y'} \right)^2} - xy' + y = 0

Option A (

{\left( {y'} \right)^2} - xy' + y = 0

) is correct because it is obtained by eliminating the arbitrary constant 'c' from the given general solution.

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y=y(t)

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