NCERT Class 12 Maths Ch 5 Misc. Ex. Solution Free Download PDf

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NCERT Class 12 Maths Ch 5 Misc. Ex. Solution

Q.1. NCERT Class 12 Maths Ch 5 Misc. Ex. Solution

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Q.1.\ \left( 3x^{2} -9x+5\right)^{9}\\
Solution:\ \\
y=\left( 3x^{2} -9x+5\right)^{9}\\
Differentiate\ w.r.t.\ x.\\
\frac{dy}{dx} =\frac{d\left( 3x^{2} -9x+5\right)^{9}}{dx}\\
=9\left( 3x^{2} -9x+5\right)^{8}\frac{d\left( 3x^{2} -9x+5\right)}{dx}\\
=9\left( 3x^{2} -9x+5\right)^{8}( 6x-9)\\
=9\left( 3x^{2} -9x+5\right)^{8} 3( 2x-3)\\
=27\left( 3x^{2} -9x+5\right)( 2x-3) \ Ans.
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Q.2. NCERT Class 12 Maths Ch 5 Misc. Ex. Solution

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Q.2.\ sin^{3} x+cos^{6} x\\
Solution:\\
y=sin^{3} x+cos^{6} x\\
\frac{dy}{dx} =\frac{d\left( sin^{3} x+cos^{6} x\right)}{dx}\\
using\ chain\ rule.\\
=3sin^{2} x\times cosx+6cos^{5} x\times ( -sinx)\\
=3sin^{2} xcosx-6cos^{5} xsinx\\
=3sinxcosx\left( sinx-2cos^{4} x\right) \ Ans.\\
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Q. 3. NCERT Class 12 Maths Ch 5 Misc. Ex. Solution

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Q.3.\ 5x^{3cos2x}\\
Solution:\\
y=5x^{3cos2x}\\
Taking\ log\ on\ both\ sides,\ we\ get\\
logy=log\left( 5x^{3cos2x}\right)\\
logy=3cos2xlog5x\\
Diff.\ w.r.t.\ x\\
\frac{1}{y}\frac{dy}{dx} =3cos2x\frac{d( log5x)}{dx} +log5x\frac{d( 3cos2x)}{dx}\\
\Longrightarrow \frac{1}{y}\frac{dy}{dx} =3cos2x\times \frac{1}{5x} \times 5+log5x\times -3sin2x\times 2\\
\Longrightarrow \frac{1}{y}\frac{dy}{dx} =\frac{3cos2x}{log5x} \ -\ 6sin2xlog5x\\
\Longrightarrow \frac{dy}{dx} =y\left[\frac{3cos2x}{x} -6sin2xlog5x\right]\\
\Longrightarrow \frac{dy}{dx} =5x^{3cos2x}\left[\frac{3cos2x}{x} -6sin2xlog5x\right] \ Ans.
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Q.4. NCERT Class 12 Maths Ch 5 Misc. Ex. Solution

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Q.4.\ sin^{-1} x\sqrt{x} \\
Solution:\\
y=sin^{-1} x\sqrt{x}\\
Diff.\ w.r.t.\ x\\
\frac{dy}{dx} =\frac{d( sin^{-1} x\sqrt{x)}}{dx}\\
=\frac{1}{\sqrt{1-\left( x\sqrt{x}\right)^{2}}} \times \frac{d\left( x\sqrt{x}\right)}{dx}\\
=\frac{1}{\sqrt{1-x^{3}}} \times \frac{3}{2} x^{\frac{3}{2} -1}\\
=\frac{3}{2}\frac{\sqrt{x}}{\sqrt{1-x^{3}}} \ Ans.
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Q.5. NCERT Class 12 Maths Ch 5 Misc. Ex. Solution

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Q.5.\ \frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}} ,\ -2< x< 2\\
Solution:\\
y=\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}\\
Differenetiaate\ w.r.t.x\\
\frac{dy}{dx} =\frac{\sqrt{2x+7}\frac{d\left( cos^{-1}\frac{x}{2}\right)}{dx} -cos^{-1}\frac{x}{2}\frac{d\left(\sqrt{2x+7}\right)}{dx}}{2x+7}\\
=\frac{\sqrt{2x+7} \times \frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} \times \frac{1}{2} -cos^{1}\frac{x}{2} \times \frac{1}{2\sqrt{2x+7}} \times 2}{2x+7}\\
=\frac{-\frac{\sqrt{2x+7}}{\sqrt{4-x^{2}}} \times 2\times \frac{1}{2} -\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}}{2x+7}\\
=\frac{-\frac{\sqrt{2x+7}}{\sqrt{4-x^{2}}} -\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}}{2x+7} \\ Ans.
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Q.6. NCERT Class 12 Maths Ch 5 Misc. Ex. Solution

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Q.6.\ \ Cot^{-1}\left[\frac{\sqrt{1+sinx} +\sqrt{1-sinx}}{\sqrt{1+sinx} -\sqrt{1-sinx}}\right] ,\ 0< x< \frac{\pi }{2}\\
Solution:\\
y=cot^{-1}\left[\frac{\sqrt{1+sinx} +\sqrt{1-sinx}}{\sqrt{1+sinx} -\sqrt{1-sinx}}\right]\\
We\ have\ 1+sinx\ =\ cos^{2}\frac{x}{2} +sin^{2}\frac{x}{2} +2cos\frac{x}{2} sin\frac{x}{2}\\
So,\ \sqrt{1+sinx} =cos\frac{x}{2} +sin\frac{x}{2}\\
\sqrt{1-sinx} =cos\frac{x}{2} -sin\frac{x}{2}\\
Substituting\ these\ values.\\
y=cot^{-1}\left[\frac{cos\frac{x}{2} +sin\frac{x}{2} \ +\ cos\frac{x}{2} -sin\frac{x}{2}}{cos\frac{x}{2} +sin\frac{x}{2} -cos\frac{x}{2} +sin\frac{x}{2}}\right]\\
=cot^{-1}\left[\frac{2cos\frac{x}{2}}{2sin\frac{x}{2}}\right]\\
=cot^{-1}\left( cot\frac{x}{2}\right) =\frac{x}{2}\\
y=\frac{x}{2}\\
\frac{dy}{dx} =\frac{1}{2}\\
Ans.\ \frac{1}{2}
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Q.7. NCERT Class 12 Maths Ch 5 Misc. Ex. Solution

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Q.7.\ ( logx)^{logx} ,\ x >1\\
Solution:\\
y=( logx)^{logx}\\
Taking\ log\ on\ both\ sides,\\
logy=logx( log( logx))\\
Differentiating\ w.r.t\ x,\ we\ get\\
\frac{1}{y}\frac{dy}{dx} =\frac{d( logxlog( logx)}{dx}\\
\frac{1}{y}\frac{dy}{dx} =logx\times \frac{d( log( logx))}{dx} +log( logx) \times \frac{d( logx)}{dx}\\
\Longrightarrow \frac{1}{y}\frac{dy}{dx} =logx\times \frac{1}{logx} \times \frac{1}{x} +log( logx) \times \frac{1}{x}\\
\Longrightarrow \frac{dy}{dx} =y\left[\frac{1}{x} +\frac{log( logx)}{x}\right]\\
\Longrightarrow \frac{dy}{dx} =( logx)^{logx}\left[\frac{1+log( logx)}{x}\right] \ Ans.
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Q.8.NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.8.\ cos( acosx+bsinx) ,\ for\ some\ constant\ a\ and\ b.\\
Solution:\\
y=cos( acosx+bsinx)\\
Differentiating\\
\frac{dy}{dx} =-sin( acoscx+bsinx) \times \frac{d( acosx+bsinx)}{dx}\\
=-sin( acosx+bsinx) \times ( -asinx+bcosx)\\
=( asinx-bcosx) .sin( acosx+bsinx\ Ans.\\
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Q.9. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.9.\ \ ( sinx-cosx)^{sinx-cosx}\\
Solution:\\
y=( sinx-cosx)^{sinx-cosx}\\
Taking\ log\ on\ both\ sides,\\
logy=( sinx-cosx) log( sinx-cosx)\\
Differentiating\ w.r.t.\ x\\
\frac{1}{y}\frac{dy}{dx} =( sinx-cosx)\frac{d( log( sinx-cosx))}{dx} +\\
log( sinx-cosx)\frac{d( sinx-cosx)}{dx}\\
\Longrightarrow \frac{1}{y}\frac{dy}{dx} =( sinx-cosx)\frac{1}{sinx-cosx}( cosx+sinx)\\
+log( sinx-cosx) \times ( cosx+sinx)\\
\Longrightarrow \frac{dy}{dx} =y[( cosx+sinx) +log( sinx-cosx) \times ( cosx+sinx)]\\
\Longrightarrow \frac{dy}{dx} =( sinx-cosx)^{( sinx-cosx)}( cosx+sinx)[ 1+log( sinx-cosx)]\\
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Q.10. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.\ 10.\ x^{x} +x^{a} +a^{x} +a^{a} ,\ for\ some\ fixed\ a >0\ and\ x >0\\
Solution:\\
y=x^{x} +x^{a} +a^{x} +a^{a}\\
Let\ u=x^{x}\\
Taking\ log\ on\ both\ sides,\ we\ get\\
logu=log\left( x^{x}\right)\\
\Longrightarrow log\ u=xlogx\\
Differentiating\ w.r.t.\ x\\
\frac{1}{u}\frac{du}{dx} =x\left(\frac{d( logx)}{dx}\right) +logx\left(\frac{dx}{dx}\right)\\
\frac{du}{dx} =u\left[\left( x\times \frac{1}{x}\right) +logx( 1)\right]\\
\frac{du}{dx} =x^{x}[ 1+logx]\\
Also,\ \frac{d\left( a^{x}\right)}{dx} =a^{x} loga\\
\frac{d\left( x^{a}\right)}{dx} =ax^{a-1}\\
Plug\ in\ all\ these\ values\\
\frac{dy}{dx} =x^{x}[ 1+logx] +ax^{a-1} +a^{x} loga+0\ Ans.
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Q. 11. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.11.\ x^{x^{2} -3} +( x-3){^{x}}^{2} ,\ for\ x >3\\
Solution:\\
y=x^{x^{2} -3} +( x-3){^{x}}^{2}\\
y=u+v\\
\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx} \ ……………………….( 1)\\
u=x^{x^{2} -3}\\
logu=\left( x^{2} -3\right) logx\\
Differentiating\ w.r.t.x\\
\frac{\ 1}{u}\frac{du}{dx} =\left( x^{2} -3\right) \times \frac{1}{x} +logx\times ( 2x)\\
\frac{du}{dx} =u\left[\frac{x^{2} -3+2x^{2} logx}{x}\right]\\
\frac{du}{dx} =x^{x^{2} -3}\left[\frac{x^{2}( 1+2logx) -3}{x}\right]\\
v=( x-3)^{x^{2}}\\
logv=x^{2} log( x-3)\\
Differentiating\ w.r.t.\ x\\
\frac{1}{v}\frac{dv}{dx} =x^{2} \times \left(\frac{1}{x-3}\right) +log( x-3) \times 2x\\
\frac{dv}{dx} =v\left[\frac{x^{2}}{x-3} +2xlog( x-3)\right]\\
\frac{dv}{dx} =( x-3){^{x}}^{2}\left[\frac{x^{2}}{x-3} +2xlog( x-3)\right]\\
Put\ \frac{du}{dx} \ and\ \frac{dv}{dx\ } \ in\ equation\ ( 10\\
\frac{dy}{dx} =\ x^{x^{2} -3}\left[\frac{x^{2}( 1+2logx) -3}{x}\right] +( x-3){^{x}}^{2}\left[\frac{x^{2}}{x-3} +2xlog( x-3)\right] Ans.\\
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Q. 12. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.12.\ Find\ \frac{dy}{dx} ,\ if\ y=12( 1-cost) ,\ x=10( t-sint) ,\ -\frac{\pi }{2} < t< \frac{\pi }{2}\\
Solution:\\
y=12( 1-cost)\\
Differentiating\ w.r.t.\ t,\\
\frac{dy}{dt} =12( 0+sint) =12sint\\
x=10( 1-sint)\\
Differentiating\ w.r.t.\ t\\
\frac{dx}{dt} =10( 1-cost)\\
\Longrightarrow \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{12sint}{10( 1-cost)}\\
=\frac{6}{5}\frac{2sin\frac{t}{2} cos\frac{t}{2}}{2sin^{2}\frac{t}{2}} =\frac{6}{5} cot\frac{t}{2} \\ Ans.
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Q. 13. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.13.\ Find\ \frac{dy}{dx} ,\ if\ y=sin^{-1} x+sin^{-1}\sqrt{1-x^{2}} ,\ 0< x< 1\\
Solution:\\
y=sin^{-1} x+sin^{-1}\sqrt{1-x^{2}}\\
Differentiating\ w.r.t.\ x\\
\frac{dy}{dx} =\frac{1}{\sqrt{1-x^{2}}} +\frac{1}{\sqrt{1-\left(\sqrt{1-x^{2}}\right)^{2}}} \times \frac{1}{2\sqrt{1-x^{2}}} \times -2x\\
\frac{dy}{dx} =\frac{1}{\sqrt{1-x^{2}}} -\frac{2x}{\sqrt{1-1+x^{2}}\sqrt{1-x^{2}}}\\
\frac{dy}{dx} =\frac{1}{\sqrt{1-x^{2}}} -\frac{x}{x\sqrt{1-x^{2}}}\\
\frac{dy}{dx} =\ \frac{1}{\sqrt{1-x^{2}}} -\frac{1}{\sqrt{1-x^{2}}} =0\ Ans.
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Q. 14. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.14.\ If\ x\sqrt{1+y} \ +y\sqrt{1+x} =0\ Prove\ that:\ \frac{dy}{dx} =-\frac{1}{( 1+x)^{2}}\\
Solution:\\
We\ have,\\
x\sqrt{1+y} +y\sqrt{1+x} \ =0\\
\Longrightarrow x\sqrt{1+y} =-y\sqrt{1+x}\\
Squaring\ both\ sides,\\
x^{2}( 1+y) =y^{2}( 1+x)\\
\Longrightarrow x^{2} +x^{2} y=y^{2} +y^{2} x\\
\Longrightarrow x^{2} -y^{2} +x^{2} y-y^{2} x=0\\
\Longrightarrow ( x-y)( x+y) +xy( x-y) =0\\
\Longrightarrow ( x-y)( x+y+xy) =0\\
\Longrightarrow x-y=0\ or,\ x+y+xy=0\\
But\ x\neq y,\ So,\ x+y+xy=0\\
\Longrightarrow y( 1+x) =-x\\
\Longrightarrow y=-\frac{x}{1+x}\\
\Longrightarrow \frac{dy}{dx} =-\frac{1}{( 1+x)^{2}} ,\ [ Using\ quotient\ rule]\\
Proved.
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Q. 15. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

Q.15. If ( (x-a)^2+(y-b)^2=c^2 ), for some ( c>0 ), prove that
( [1+(dy/dx)^2]^{3/2}/(d^2y/dx^2) ) is a constant independent of ( a ) and ( b ).

Solution:

Given: ( (x-a)^2+(y-b)^2=c^2 ) …(i)

Differentiating with respect to ( x ),
( 2(x-a)+2(y-b)(dy/dx)=0 )

( (x-a)+(y-b)(dy/dx)=0 ) …(ii)

Again differentiating with respect to ( x ),
( 1+(dy/dx)^2+(y-b)(d^2y/dx^2)=0 ) …(iii)

From (iii),
( (y-b)=-[1+(dy/dx)^2]/(d^2y/dx^2) )

Substituting this in (ii),
( (x-a)-1+(dy/dx)^2/(d^2y/dx^2)=0 )

So,
( (x-a)=1+(dy/dx)^2/(d^2y/dx^2) )

Substituting values of ( (x-a) ) and ( (y-b) ) in (i),
( (1+(dy/dx)^2/(d^2y/dx^2))^2 + ([1+(dy/dx)^2]/(d^2y/dx^2))^2=c^2 )

This gives,
( [1+(dy/dx)^2]^3/(d^2y/dx^2)^2=c^2 )

Taking square root,
( [1+(dy/dx)^2]^{3/2}/(d^2y/dx^2)=c )

Hence proved.

Q. 16. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.16.\ cosy=xcos( a+y) \ with\ cosa\neq 1,\ Prove\ that\ \frac{dy}{dx} =\frac{cos^{2}( a+y)}{sina}\\
Solution:\\
Given:\ cosy=xcos( a+y)\\
\Longrightarrow x=\frac{cosy}{cos( a+y)}\\
Differentiating\ w.r.t.\ y,\\
\frac{dx}{dy} =\frac{cos( a+y) \times \frac{d( cosy)}{dy} -cosy\frac{d( cos( a+y))}{dy}}{cos^{2}( a+y)}\\
\frac{dx}{dy} =\frac{cos( a+y) .( -siny) -cosy( -sin( a+y)}{cos^{2}( a+y)}\\
\frac{dx}{dy} =\frac{-cos( a+y) siny+sin( a+y) cosy}{cos^{2}( a+y)}\\
\frac{dx}{dy} =\frac{sin( a+y-y)}{cos^{2}( a+y)} =\frac{sina}{cos^{2}( a+y)}\\
\Longrightarrow \frac{dy}{dx} =\frac{cos^{2}( a+y)}{sina} \ Proved.
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Q. 17. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.17.\ \ If\ x=a( cost+tsint) \ and\ y=a( sint-tcost) ,\ find\ \frac{d^{2} y}{dx^{2}} .\\
Solution:\\
x=a( cost+tsint)\\
Differentiating\ w.r.t.\ t\\
\frac{dx}{dt} =a( -sint+tcost+sint( 1))\\
\Longrightarrow \frac{dx}{dt} =atcost\ \ \ ……….\ ( i)\\
y=a( sint-tcost)\\
Differentiating\ w.r.t.\ t\\
\frac{dy}{dt} =a( cost-( t( -sint) +cost( 1)))\\
\Longrightarrow \frac{dy}{dt} =a( cost+tsint-cost)\\
\Longrightarrow \frac{dy}{dt} =atsint\\
Now,\ \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{atsint}{atcost} =tant\\
\Longrightarrow \frac{dy}{dx} =tant\\
Differentiating\ w.r.t.\ x\\
\frac{d^{2} y}{dx^{2}} =sec^{2} t.\frac{dt}{dx}\\
From\ ( i)\\
\frac{d^{2} y}{dx^{2}} =sec^{2} t\times \frac{1}{atcost}\\
\Longrightarrow \frac{d^{2} y}{dx^{2}} =\frac{sec^{3} t}{at} \ Ans.
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Q. 22. NCERT Class Xii maths Chapter 5 Miscellaneous Exercise Solution

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Q.22.\ If\ y=e^{acos^{-1} x} ‘-1< x< 1,\ show\ that\ \left( 1-x^{2}\right)\frac{d^{2} y}{dx^{2}} -x\frac{dy}{dx} -a^{2} y=0\\
Solution:\\
y=e^{acos^{-1} x} \ ………..\ \ ( i)\\
Differentiating\ w.r.t\ x\\
\frac{dy}{dx} =e^{acos^{-1} x} .\frac{d\left( acos^{-1} x\right)}{dx}\\
\Longrightarrow \frac{dy}{dx} =e^{acos^{-1} x} .\left(\frac{-a}{\sqrt{1-x^{2}}}\right) =\frac{-ay}{\sqrt{1-x^{2}}} \ [ From\ ( i)] \ \ \ ……\ ( ii)\\
Again\ differentiating\ w.r.t.\ x,\ we\ get\\
\frac{d^{2} y}{dx^{2}} =\ \frac{\sqrt{1-x^{2}}\left(\frac{d( -ay)}{dx}\right) -( -ay)\frac{d\left(\sqrt{1-x^{2}}\right)}{dx}}{\left(\sqrt{1-x^{2}}\right)^{2}}\\
=\frac{\sqrt{1-x^{2}}\left( -a\frac{dy}{dx}\right) +ay\frac{1}{2\sqrt{1-x^{2}}} .( -2x)}{1-x^{2}}\\
Substituting\ \frac{dy}{dx} \ from\ ( ii)\\
\Longrightarrow \left( 1-x^{2}\right)\frac{d^{2} y}{dx^{2}} =\sqrt{1-x^{2}}\left( -a\times \frac{-ay}{\sqrt{1-x^{2}}}\right) +x\frac{dy}{dx}\\
\Longrightarrow \left( 1-x^{2}\right)\frac{d^{2} y}{dx^{2}} =a^{2} y+x\frac{dy}{dx}\\
\Longrightarrow \left( 1-x^{2}\right)\frac{d^{2} y}{dx^{2}} -x\frac{dy}{dx} -a^{2} y=0\ Proved.
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