Let y=y(x) be the solution of the differential equation (x+1) y^{'}-y=e^{3 x}(x+1)^2, with y(0)=\frac{1}{3}. Then, the point x=-\frac{4}{3} for the curve y=y(x) is : [25-Jun-2022-Shift-1]

Correct Answer is : a point of local minima

JEE Main 25 June 2022 Shift 1 Solved Paper

Section: Mathematics

Question : 15 of 90

Marks: +1, -0

Let y=y(x) be the solution of the differential equation (x+1)y′−y=e3x(x+1)2, with y(0)=‌

. Then, the point x=−‌

for the curve y=y(x) is :

[25-Jun-2022-Shift-1]

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