9:null 17:T523,undefined Given: y^1/m + y^-1/m = 2x Find: (x^2 - 1)d^2y/dx^2 + xdy/dx

Let u = y^1/m Then y^-1/m = 1/u The equation becomes: u + 1/u = 2x

Multiply both sides by u: u^2 + 1 = 2xu u^2 - 2xu + 1 = 0 Using the quadratic formula: u = 2x \± sqrt(4x^2 - 4)2 u = 2x \± 2sqrt(x^2-1)2 u = x \± sqrt(x^2 - 1)

Since u = y^1/m: y^1/m = x \± sqrt(x^2 - 1) Raising both sides to power m: y = (x \± sqrt(x^2 - 1))^m

Using chain ru<=: dy/dx = m(x \± sqrt(x^2 - 1))^m-1 d/dx(x \± sqrt(x^2 - 1)) d/dx(x \± sqrt(x^2 - 1)) = 1 \± xsqrt(x^2-1) = sqrt(x^2-1) \± xsqrt(x^2-1) = x \± sqrt(x^2-1)sqrt(x^2-1) Therefore: dy/dx = m(x \± sqrt(x^2 - 1))^m-1 x \± sqrt(x^2-1)sqrt(x^2-1) = m (x \± sqrt(x^2 - 1))^msqrt(x^2-1) Since y = (x \± sqrt(x^2 - 1))^m: dy/dx = mysqrt(x^2-1)

Using quotient ru<= on dy/dx = mysqrt(x^2-1): d^2y/dx^2 = mdy/dx sqrt(x^2-1) - my xsqrt(x^2-1)x^2-1 = m(x^2-1)dy/dx - mxy(x^2-1)^3/2

Let u = y^1/m Then y^-1/m = 1/u The equation becomes: u + 1/u = 2x

Multiply both sides by u: u^2 + 1 = 2xu u^2 - 2xu + 1 = 0 Using the quadratic formula: u = 2x \± sqrt(4x^2 - 4)2 u = 2x \± 2sqrt(x^2-1)2 u = x \± sqrt(x^2 - 1)

Since u = y^1/m: y^1/m = x \± sqrt(x^2 - 1) Raising both sides to power m: y = (x \± sqrt(x^2 - 1))^m

Using quotient ru<= on dy/dx = mysqrt(x^2-1): d^2y/dx^2 = mdy/dx sqrt(x^2-1) - my xsqrt(x^2-1)x^2-1 = m(x^2-1)dy/dx - mxy(x^2-1)^3/2

(x^2-1)d^2y/dx^2 = m(x^2-1)dy/dx - mxysqrt(x^2-1) (x^2-1)d^2y/dx^2 + xdy/dx = m(x^2-1)dy/dx - mxysqrt(x^2-1) + xdy/dx Substituting dy/dx = mysqrt(x^2-1): = m(x^2-1) mysqrt(x^2-1) - mxysqrt(x^2-1) + x mysqrt(x^2-1) = m^2ysqrt(x^2-1) - mxysqrt(x^2-1) + mxysqrt(x^2-1) = m^2ysqrt(x^2-1) - mxy + mxysqrt(x^2-1) = m^2ysqrt(x^2-1)sqrt(x^2-1) = m^2y Therefore, (x^2 - 1)d^2y/dx^2 + xdy/dx = m^2y10:[["$","title","0",{"children":"CUET Mathematics - If y^1/m + y^-1/m = 2x, then the value of (x^2 - 1)d^2y/dx^2 + xdy/dx is: | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"$17"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"CUETMAT,MAT,Calculus,Continuity & Differentiability"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - If y^1/m + y^-1/m = 2x, then the value of (x^2 - 1)d^2y/dx^2 + xdy/dx is: | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"$18"}],"$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20","$L21","$L22","$L23","$L24","$L25","$L26","$L27","$L28","$L29","$L2a"]

CUET Mathematics 2025 19 May Shift 1 PYQs

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