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Differential Equations question

2021 · 17 Mar · Shift 1 · Q26
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  5. /2021 · 17 Mar · Shift 1 · Q26

Differential Equations question

2021 · 17 Mar · Shift 1 · Q26

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Which of the following is true for y(x) that satisfies the differential equation dydx{{dy} \over {dx}}dxdy​= xy −-− 1 + x −-− y; y(0) = 0 :
  1. A
    y(1) = 1
  2. B
    y(1) = e −12−-{1 \over 2}-−21​− 1
  3. C
    y(1) = e 12−{1 \over 2}-21​− e −12-{1 \over 2}−21​
  4. D
    y(1) = e 12−{1 \over 2}-21​− 1
View written solutionFree

Correct answer: B

  1. Rewrite the differential equation

Given dydx=xy−1+x−y,y(0)=0.\frac{dy}{dx}=xy-1+x-y, \qquad y(0)=0.dxdy​=xy−1+x−y,y(0)=0.

Group the terms involving yyy: dydx=y(x−1)+(x−1).\frac{dy}{dx}=y(x-1)+(x-1).dxdy​=y(x−1)+(x−1). So, dydx=(x−1)(y+1).\frac{dy}{dx}=(x-1)(y+1).dxdy​=(x−1)(y+1).

  1. Separate variables

dyy+1=(x−1) dx.\frac{dy}{y+1}=(x-1)\,dx.y+1dy​=(x−1)dx.

  1. Integrate both sides

∫dyy+1=∫(x−1) dx\int \frac{dy}{y+1}=\int (x-1)\,dx∫y+1dy​=∫(x−1)dx

which gives ln⁡∣y+1∣=x22−x+C.\ln|y+1|=\frac{x^2}{2}-x+C.ln∣y+1∣=2x2​−x+C.

  1. Use the initial condition

Given y(0)=0y(0)=0y(0)=0: ln⁡∣0+1∣=022−0+C\ln|0+1|=\frac{0^2}{2}-0+Cln∣0+1∣=202​−0+C 0=C.0=C.0=C.

Hence, ln⁡(y+1)=x22−x.\ln(y+1)=\frac{x^2}{2}-x.ln(y+1)=2x2​−x. Since y(0)=0y(0)=0y(0)=0, we have y+1>0y+1>0y+1>0 near the solution, so y+1=ex22−x.y+1=e^{\frac{x^2}{2}-x}.y+1=e2x2​−x. Thus, y=ex22−x−1.y=e^{\frac{x^2}{2}-x}-1.y=e2x2​−x−1.

  1. Find y(1)y(1)y(1)

Substitute x=1x=1x=1: y(1)=e12−1−1=e−12−1.y(1)=e^{\frac{1}{2}-1}-1=e^{-\frac12}-1.y(1)=e21​−1−1=e−21​−1.

  1. Match with the options

Option B is e−12−1,e^{-\frac12}-1,e−21​−1, which matches exactly.

Therefore, the correct option is B.

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