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Differential Equations question

2015 · Shift 1 · Q29
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  5. /2015 · Shift 1 · Q29

Differential Equations question

2015 · Shift 1 · Q29

JEE AdvancedMathematicsDifferential EquationsMultiple correct+4 / −2
Consider the family of all circles whose centres lie on the straight line y=x,y=x,y=x, If this family of circle is represented by the differential equation Py′′+Qy′+1=0,Py'' + Qy' + 1 = 0,Py′′+Qy′+1=0, where P,QP, QP,Q are functions of x,yx,yx,y and y′y'y′ (here   y′=dydx,y′′=d2ydx2)\left( {here\,\,\,y' = {{dy} \over {dx}},y'' = {{{d^2}y} \over {d{x^2}}}} \right)(herey′=dxdy​,y′′=dx2d2y​) then which of the following statements is (are) true?
  1. A
    P=y+xP = y + xP=y+x
  2. B
     P=y−x\,P = y - xP=y−x
  3. C
     P+Q=1−x+y+y′+(y′)2\,P + Q = 1 - x + y + y' + {\left( {y'} \right)^2}P+Q=1−x+y+y′+(y′)2
  4. D
     P−Q=1−x+y−y′−(y′)2\,P - Q = 1 - x + y - y' - {\left( {y'} \right)^2}P−Q=1−x+y−y′−(y′)2
View written solutionFree

Correct answer: B, C

Step-by-step Derivation:

  1. Formulate the equation of the family of circles. Let the center of a circle be (h,k)(h, k)(h,k) and its radius be rrr. The equation of the circle is (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2(x−h)2+(y−k)2=r2. The problem states that the centers lie on the straight line y=xy=xy=x. Therefore, the coordinates of the center are (h,h)(h, h)(h,h). The equation for the family of circles is: (x−h)2+(y−h)2=r2(x-h)^2 + (y-h)^2 = r^2(x−h)2+(y−h)2=r2 This equation has two arbitrary constants, hhh and rrr. To find the differential equation for this family, we need to differentiate this equation twice with respect to xxx to eliminate these constants.

  2. First Differentiation. Differentiating the equation with respect to xxx, we get: 2(x−h)(1)+2(y−h)dydx=02(x-h)(1) + 2(y-h) \frac{dy}{dx} = 02(x−h)(1)+2(y−h)dxdy​=0 Let y′=dydxy' = \frac{dy}{dx}y′=dxdy​. The equation simplifies to: (x−h)+(y−h)y′=0(1)(x-h) + (y-h)y' = 0 \quad \quad (1)(x−h)+(y−h)y′=0(1) This step has eliminated the constant rrr. Now we need to eliminate hhh.

  3. Second Differentiation. Differentiating Equation (1) with respect to xxx: 1+[ddx(y−h)⋅y′+(y−h)⋅ddx(y′)]=01 + \left[ \frac{d}{dx}(y-h) \cdot y' + (y-h) \cdot \frac{d}{dx}(y') \right] = 01+[dxd​(y−h)⋅y′+(y−h)⋅dxd​(y′)]=0 1+[y′⋅y′+(y−h)y′′]=01 + [y' \cdot y' + (y-h)y''] = 01+[y′⋅y′+(y−h)y′′]=0 Let y′′=d2ydx2y'' = \frac{d^2y}{dx^2}y′′=dx2d2y​. The equation becomes: 1+(y′)2+(y−h)y′′=0(2)1 + (y')^2 + (y-h)y'' = 0 \quad \quad (2)1+(y′)2+(y−h)y′′=0(2)

  4. Eliminate the constant h. From Equation (2), we can express (y−h)(y-h)(y−h) in terms of y′y'y′ and y′′y''y′′: (y−h)y′′=−(1+(y′)2)(y-h)y'' = - (1 + (y')^2)(y−h)y′′=−(1+(y′)2) y−h=−1+(y′)2y′′(3)y-h = -\frac{1+(y')^2}{y''} \quad \quad (3)y−h=−y′′1+(y′)2​(3) From Equation (1), we have (x−h)=−(y−h)y′(x-h) = -(y-h)y'(x−h)=−(y−h)y′. Substituting the expression for (y−h)(y-h)(y−h) from (3): x−h=−(−1+(y′)2y′′)y′=y′(1+(y′)2)y′′(4)x-h = -\left(-\frac{1+(y')^2}{y''}\right)y' = \frac{y'(1+(y')^2)}{y''} \quad \quad (4)x−h=−(−y′′1+(y′)2​)y′=y′′y′(1+(y′)2)​(4) Now, we can find y−xy-xy−x by subtracting Equation (4) from Equation (3): (y−h)−(x−h)=y−x(y-h) - (x-h) = y-x(y−h)−(x−h)=y−x y−x=−1+(y′)2y′′−y′(1+(y′)2)y′′y-x = -\frac{1+(y')^2}{y''} - \frac{y'(1+(y')^2)}{y''}y−x=−y′′1+(y′)2​−y′′y′(1+(y′)2)​ y−x=−(1+(y′)2)(1+y′)y′′y-x = -\frac{(1+(y')^2)(1+y')}{y''}y−x=−y′′(1+(y′)2)(1+y′)​ Multiplying both sides by y′′y''y′′: (y−x)y′′=−(1+(y′)2)(1+y′)(y-x)y'' = -(1+(y')^2)(1+y')(y−x)y′′=−(1+(y′)2)(1+y′) (y−x)y′′=−(1+y′+(y′)2+(y′)3)(y-x)y'' = -(1 + y' + (y')^2 + (y')^3)(y−x)y′′=−(1+y′+(y′)2+(y′)3) Rearranging the terms to one side, we get the differential equation: (y−x)y′′+1+y′+(y′)2+(y′)3=0(y-x)y'' + 1 + y' + (y')^2 + (y')^3 = 0(y−x)y′′+1+y′+(y′)2+(y′)3=0

  5. Compare with the given form. The given differential equation is Py′′+Qy′+1=0Py'' + Qy' + 1 = 0Py′′+Qy′+1=0. Let's rearrange our derived equation to match this form: (y−x)y′′+(y′+(y′)2+(y′)3)+1=0(y-x)y'' + (y' + (y')^2 + (y')^3) + 1 = 0(y−x)y′′+(y′+(y′)2+(y′)3)+1=0 We can factor out y′y'y′ from the second term: (y−x)y′′+(1+y′+(y′)2)y′+1=0(y-x)y'' + (1 + y' + (y')^2)y' + 1 = 0(y−x)y′′+(1+y′+(y′)2)y′+1=0 Comparing this with Py′′+Qy′+1=0Py'' + Qy' + 1 = 0Py′′+Qy′+1=0, we can identify PPP and QQQ: P=y−xP = y-xP=y−x Q=1+y′+(y′)2Q = 1 + y' + (y')^2Q=1+y′+(y′)2

  6. Evaluate the options. Now we check the given statements using these expressions for PPP and QQQ.

    • Option A: P=y+xP = y + xP=y+x This is incorrect. We found P=y−xP = y - xP=y−x.

    • Option B: P=y−xP = y - xP=y−x This is correct.

    • Option C: P+Q=1−x+y+y′+(y′)2P + Q = 1 - x + y + y' + (y')^2P+Q=1−x+y+y′+(y′)2 Let's compute P+QP+QP+Q: P+Q=(y−x)+(1+y′+(y′)2)P + Q = (y-x) + (1 + y' + (y')^2)P+Q=(y−x)+(1+y′+(y′)2) P+Q=1−x+y+y′+(y′)2P + Q = 1 - x + y + y' + (y')^2P+Q=1−x+y+y′+(y′)2 This statement is correct.

    • Option D: P−Q=1−x+y−y′−(y′)2P - Q = 1 - x + y - y' - (y')^2P−Q=1−x+y−y′−(y′)2 Let's compute P−QP-QP−Q: P−Q=(y−x)−(1+y′+(y′)2)P - Q = (y-x) - (1 + y' + (y')^2)P−Q=(y−x)−(1+y′+(y′)2) P−Q=y−x−1−y′−(y′)2P - Q = y - x - 1 - y' - (y')^2P−Q=y−x−1−y′−(y′)2 The expression in the option is 1−x+y−y′−(y′)21 - x + y - y' - (y')^21−x+y−y′−(y′)2. The constant term is +1+1+1 in the option, while we found it to be −1-1−1. Thus, this statement is incorrect.

Conclusion:

Based on the derivation, the statements in options B and C are true.

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