Let y = y x be the solution of the differential equation x + 1 y ' - y = e 3 x x + 1 2 , with y 0 = 1 3 …

Let y=yx be the solution of the differential equation x+1y'-y=e3xx+12, with y0=13. Then, the point x=-43 for the curve y=yx is
  1. not a critical point
  2. a point of local minima
  3. a point of local maxima
  4. a point of inflection

Solution

x+1dy-ydx=e3xx+12dx

x+1dy-ydx(x+1)2=e3xdx

dyx+1=e3xdx

On integration, we get

yx+1=e3x3+C

Given y0=13

So C=0

⇒y=x+1e3x3

dydx=e3x33x+4

d2ydx2=e3x3x+5

Clearly, x=-43 is a point of local minima as d2ydx2x=-43<0

Asked in: JEE Main 2022 (25 Jun Shift 1)

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