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Differential Equations question

2022 · 25 Jun · Shift 1 · Q35
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  5. /2022 · 25 Jun · Shift 1 · Q35

Differential Equations question

2022 · 25 Jun · Shift 1 · Q35

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation (x+1)y′−y=e3x(x+1)2(x + 1)y' - y = {e^{3x}}{(x + 1)^2}(x+1)y′−y=e3x(x+1)2, with y(0)=13y(0) = {1 \over 3}y(0)=31​. Then, the point x=−43x = - {4 \over 3}x=−34​ for the curve y=y(x)y = y(x)y=y(x) is :
  1. A
    not a critical point
  2. B
    a point of local minima
  3. C
    a point of local maxima
  4. D
    a point of inflection
View written solutionFree

Correct answer: B

  1. Given differential equation
(x+1)y′−y=e3x(x+1)2,y(0)=13(x+1)y' - y = e^{3x}(x+1)^2, \qquad y(0)=\frac13(x+1)y′−y=e3x(x+1)2,y(0)=31​

We need to determine the nature of the point x=−43x=-\frac43x=−34​ on the solution curve y=y(x)y=y(x)y=y(x).

  1. Rewrite in linear form

Divide by x+1x+1x+1 (for x≠−1x\neq -1x=−1):

y′−1x+1y=e3x(x+1)y' - \frac{1}{x+1}y = e^{3x}(x+1)y′−x+11​y=e3x(x+1)

This is a linear differential equation:

y′+P(x)y=Q(x),P(x)=−1x+1y' + P(x)y = Q(x), \qquad P(x)=-\frac{1}{x+1}y′+P(x)y=Q(x),P(x)=−x+11​
  1. Find the integrating factor
IF=e∫−1x+1 dx=e−ln⁡∣x+1∣=1x+1IF = e^{\int -\frac{1}{x+1}\,dx} = e^{-\ln|x+1|} = \frac{1}{x+1}IF=e∫−x+11​dx=e−ln∣x+1∣=x+11​

So multiplying the equation by 1x+1\frac1{x+1}x+11​ gives

1x+1y′−1(x+1)2y=e3x\frac{1}{x+1}y' - \frac{1}{(x+1)^2}y = e^{3x}x+11​y′−(x+1)21​y=e3x

The left side is

ddx(yx+1)\frac{d}{dx}\left(\frac{y}{x+1}\right)dxd​(x+1y​)

Hence,

ddx(yx+1)=e3x\frac{d}{dx}\left(\frac{y}{x+1}\right)=e^{3x}dxd​(x+1y​)=e3x

Integrating,

yx+1=13e3x+C\frac{y}{x+1}=\frac13 e^{3x}+Cx+1y​=31​e3x+C

Thus,

y=(x+1)(13e3x+C)y=(x+1)\left(\frac13 e^{3x}+C\right)y=(x+1)(31​e3x+C)
  1. Use the initial condition

Given y(0)=13y(0)=\frac13y(0)=31​:

13=(0+1)(13e0+C)=13+C\frac13=(0+1)\left(\frac13 e^0 + C\right)=\frac13 + C31​=(0+1)(31​e0+C)=31​+C

So,

C=0C=0C=0

Therefore the solution is

y=13(x+1)e3x\boxed{y=\frac13 (x+1)e^{3x}}y=31​(x+1)e3x​
  1. Find critical points

Differentiate:

y=13(x+1)e3xy=\frac13 (x+1)e^{3x}y=31​(x+1)e3x

Using product rule,

y′=13[e3x+3(x+1)e3x]y' = \frac13\left[e^{3x} + 3(x+1)e^{3x}\right]y′=31​[e3x+3(x+1)e3x] y′=13e3x(3x+4)y' = \frac13 e^{3x}(3x+4)y′=31​e3x(3x+4)

Now at x=−43x=-\frac43x=−34​,

y′(−43)=13e−4(3(−43)+4)=13e−4(−4+4)=0y'\left(-\frac43\right)=\frac13 e^{-4}(3(-\frac43)+4)=\frac13 e^{-4}(-4+4)=0y′(−34​)=31​e−4(3(−34​)+4)=31​e−4(−4+4)=0

So x=−43x=-\frac43x=−34​ is a critical point.

  1. Determine whether it is max/min/inflection

Since e3x>0e^{3x}>0e3x>0 for all xxx, the sign of y′y'y′ depends only on 3x+43x+43x+4.

  • If x<−43x< -\frac43x<−34​, then 3x+4<03x+4<03x+4<0 so y′<0y'<0y′<0
  • If x>−43x> -\frac43x>−34​, then 3x+4>03x+4>03x+4>0 so y′>0y'>0y′>0

Thus the function changes from decreasing to increasing at x=−43x=-\frac43x=−34​.

Hence x=−43x=-\frac43x=−34​ is a point of local minimum.

  1. Check inflection if needed

Differentiate again:

y′=13e3x(3x+4)y' = \frac13 e^{3x}(3x+4)y′=31​e3x(3x+4) y′′=13[3e3x(3x+4)+3e3x]=e3x(3x+5)y'' = \frac13\left[3e^{3x}(3x+4)+3e^{3x}\right] = e^{3x}(3x+5)y′′=31​[3e3x(3x+4)+3e3x]=e3x(3x+5)

At x=−43x=-\frac43x=−34​,

y′′(−43)=e−4(3(−43)+5)=e−4(1)>0y''\left(-\frac43\right)=e^{-4}(3(-\frac43)+5)=e^{-4}(1)>0y′′(−34​)=e−4(3(−34​)+5)=e−4(1)>0

So this confirms a local minimum, not an inflection point.

  1. Conclusion

The correct option is:

B: a point of local minima\boxed{\text{B: a point of local minima}}B: a point of local minima​
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