(x+1)dy−ydx=e3x(x+1)2dx
(x+1)2(x+1)dy−ydx=e3xdx
d(x+1y)=e3xdx
On integration, we get
x+1y=3e3x+C
Given y(0)=31
So C=0
⇒y=3(x+1)e3x
dxdy=3e3x(3x+4)
dx2d2y=e3x(3x+5)
Clearly, x=3−4 is a point of local minima as (dx2d2y)x=3−4<0
JEE Main 2022 — Mathematics Calculus
Let y=y(x) be the solution of the differential equation (x+1)y′−y=e3x(x+1)2, with y(0)=31. Then, the point x=−34 for the curve y=y(x) is
Held on 25 Jun 2022 · Verified 6 Jul 2026.
not a critical point
a point of local minima
a point of local maxima
a point of inflection
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