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Question: Match the columns: List 1| List 2 ---|--- \[{e^y} + xy = e\]| \[y''\left( 0 \right) = \dfrac...

Match the columns:

List 1List 2
ey+xy=e{e^y} + xy = ey′′(0)=(1−ey(0))2(1+ey(0))3y''\left( 0 \right) = \dfrac{{{{\left( {1 - {e^{y\left( 0 \right)}}} \right)}^2}}}{{{{\left( {1 + {e^{y\left( 0 \right)}}} \right)}^3}}}
xy=ex−y{x^y} = {e^{x - y}}y′(e)=14y'\left( e \right) = \dfrac{1}{4}
x+y=ex−yx + y = {e^{x - y}}y′′(0)=1e2y''\left( 0 \right) = \dfrac{1}{{{e^2}}}
ex−ey=y−x{e^x} - {e^y} = y - xy′′(0)=4y(0)(1+y(0))3y''\left( 0 \right) = \dfrac{{4y\left( 0 \right)}}{{{{\left( {1 + y\left( 0 \right)} \right)}^3}}}
Explanation

Solution

Here, we will take each equation from list 1 and try to differentiate it once or twice. After differentiating we will find the value at x=0x = 0 or at x=ex = e and try to obtain the results as given in list2. In some cases, we will use the properties of logarithms.Also, we can write, d2ydx2\dfrac{{{d^2}y}}{{d{x^2}}} as y′′(x)y''\left( x \right). We will also use the product rule, division rule along with the chain rule of differentiation.

Complete step by step answer:
First, we will take
ey+xy=e{e^y} + xy = e
Differentiating both the sides with respect to xx, we get
⇒ddx(ey+xy)=ddx(e)\Rightarrow \dfrac{d}{{dx}}\left( {{e^y} + xy} \right) = \dfrac{d}{{dx}}\left( e \right)
Using the chain rule of differentiation, we get
⇒eydydx+xdydx+y=0\Rightarrow {e^y}\dfrac{{dy}}{{dx}} + x\dfrac{{dy}}{{dx}} + y = 0
Since ee is a constant so its differentiation is zero.

Now we will differentiate it again with respect to xx. So, we get;
⇒ddx(eydydx+xdydx+y)=ddx(0)\Rightarrow \dfrac{d}{{dx}}\left( {{e^y}\dfrac{{dy}}{{dx}} + x\dfrac{{dy}}{{dx}} + y} \right) = \dfrac{d}{{dx}}\left( 0 \right)
Again, using the chain rule of differentiation, we get
⇒ey(dydx)2+eyd2ydx2+xd2ydx2+dydx+dydx=0\Rightarrow {e^y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} + {e^y}\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} + \dfrac{{dy}}{{dx}} = 0
Now we will take d2ydx2\dfrac{{{d^2}y}}{{d{x^2}}} common from eyd2ydx2+xd2ydx2{e^y}\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{{d^2}y}}{{d{x^2}}} and on simplification, we get
⇒ey(dydx)2+(ey+x)d2ydx2+2dydx=0\Rightarrow {e^y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \left( {{e^y} + x} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + 2\dfrac{{dy}}{{dx}} = 0
On shifting we get,
⇒(ey+x)d2ydx2=−2dydx−ey(dydx)2\Rightarrow \left( {{e^y} + x} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = - 2\dfrac{{dy}}{{dx}} - {e^y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2}

Dividing both the sides by (ey+x)\left( {{e^y} + x} \right), we get;
⇒d2ydx2=−2(dydx)−ey(dydx)2(ey+x)−−−−(1)\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 2\left( {\dfrac{{dy}}{{dx}}} \right) - {e^y}{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}}{{\left( {{e^y} + x} \right)}} - - - - (1)
Now to find its value at x=0x = 0 we have to first find the value of (dydx)x=0{\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} and the value of ey  at x=0{e^{y{\text{ }}}}{\text{ at }}x = 0.
So, for this we will again take the given equation,
ey+xy=e{e^y} + xy = e
Now we will put x=0x = 0 in this equation, at x=0x = 0 we can write yy as y(0)y\left( 0 \right) because yy is a function of xx.
⇒ey(0)+0×y=e\Rightarrow {e^{y\left( 0 \right)}} + 0 \times y = e
On simplification,
⇒ey(0)=e−−−−(2)\Rightarrow {e^{y\left( 0 \right)}} = e - - - - (2)
Now we will equate the powers of both sides because the bases are the same. So, we get;
⇒y(0)=1−−−−(3)\Rightarrow y\left( 0 \right) = 1 - - - - (3)

Above we get,
⇒eydydx+xdydx+y=0\Rightarrow {e^y}\dfrac{{dy}}{{dx}} + x\dfrac{{dy}}{{dx}} + y = 0
Now we will take dydx\dfrac{{dy}}{{dx}} as common, we get
⇒(ey+x)dydx+y=0\Rightarrow \left( {{e^y} + x} \right)\dfrac{{dy}}{{dx}} + y = 0
On shifting the terms, we get,
⇒(ey+x)dydx=−y\Rightarrow \left( {{e^y} + x} \right)\dfrac{{dy}}{{dx}} = - y
Dividing both the sides by (ey+x)\left( {{e^y} + x} \right), we get
⇒dydx=−y(ey+x)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - y}}{{\left( {{e^y} + x} \right)}}
Now we will find its value at x=0x = 0. Again, at x=0x = 0 we can write yy as y(0)y\left( 0 \right) because yy is a function of xx.
⇒(dydx)x=0=−y(0)(ey(0)+0)\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = \dfrac{{ - y\left( 0 \right)}}{{\left( {{e^{y\left( 0 \right)}} + 0} \right)}}
Now we will use equation (2),(3)(2),(3) and put the values from these equations into the above equation. So, we get;
⇒(dydx)x=0=−1e−−−−(4)\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = \dfrac{{ - 1}}{e} - - - - (4)

Now we will evaluate equation (1)(1) at x=0x = 0.
⇒(d2ydx2)x=0=−2(dydx)x=0−ey(0)(dydx)x=02ey(0)+0\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{ - 2{{\left( {\dfrac{{dy}}{{dx}}} \right)}_{x = 0}} - {e^{y\left( 0 \right)}}\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}^2}}{{{e^{y\left( 0 \right)}} + 0}}
Now we will use equation (4)(4) and (2)(2) in equation (1)(1).
⇒(d2ydx2)x=0=−2(−1e)−e(−1e)2e\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{ - 2\left( {\dfrac{{ - 1}}{e}} \right) - e{{\left( {\dfrac{{ - 1}}{e}} \right)}^2}}}{e}
On simplification,
⇒(d2ydx2)x=0=−2(−1e)−e(1e2)e\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{ - 2\left( {\dfrac{{ - 1}}{e}} \right) - e\left( {\dfrac{1}{{{e^2}}}} \right)}}{e}
On further simplification we have,
⇒y′′(0)=2e−1ee\Rightarrow y''\left( 0 \right) = \dfrac{{\dfrac{2}{e} - \dfrac{1}{e}}}{e}
We will subtract the numerator. So,
⇒y′′(0)=1ee\Rightarrow y''\left( 0 \right) = \dfrac{{\dfrac{1}{e}}}{e}
On simplification,
⇒y′′(0)=1e2\Rightarrow y''\left( 0 \right) = \dfrac{1}{{{e^2}}}

So, the first option of list 1 matches with the third option of list 2.

Now we will take the second question from list 1.
xy=ex−y{x^y} = {e^{x - y}}
We will take logarithms of both sides
⇒ln⁡xy=ln⁡ex−y\Rightarrow \ln {x^y} = \ln {e^{x - y}}
Differentiating both the sides, we get
⇒yln⁡x=(x−y)ln⁡e\Rightarrow y\ln x = \left( {x - y} \right)\ln e
We know ln⁡e=1\ln e = 1, so we get,
⇒yln⁡x=x−y\Rightarrow y\ln x = x - y

Now, differentiating with respect to xx,
⇒ddx(yln⁡x)=ddx(x−y)\Rightarrow \dfrac{d}{{dx}}\left( {y\ln x} \right) = \dfrac{d}{{dx}}\left( {x - y} \right)
Using the chain rule of differentiation, we get
⇒yx+dydxln⁡x=1−dydx\Rightarrow \dfrac{y}{x} + \dfrac{{dy}}{{dx}}\ln x = 1 - \dfrac{{dy}}{{dx}}
Taking dydx\dfrac{{dy}}{{dx}} from R.H.S. to L.H.S. and yx\dfrac{y}{x} from L.H.S. to R.H.S., we get
⇒dydx+dydxln⁡x=1−yx\Rightarrow \dfrac{{dy}}{{dx}} + \dfrac{{dy}}{{dx}}\ln x = 1 - \dfrac{y}{x}
Taking dydx\dfrac{{dy}}{{dx}} common from L.H.S., we get
⇒(1+ln⁡x)dydx=1−yx\Rightarrow \left( {1 + \ln x} \right)\dfrac{{dy}}{{dx}} = 1 - \dfrac{y}{x}
Dividing both the sides by (1+ln⁡x)(1 + \ln x),
⇒dydx=1−yx1+ln⁡x−−−−(5)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{1 - \dfrac{y}{x}}}{{1 + \ln x}} - - - - (5)
Now we will find the value of yy at x=ex = e.
We have;
xy=ex−y{x^y} = {e^{x - y}}
Now we will put x=ex = e. So, we get;
⇒ey=ee−y\Rightarrow {e^y} = {e^{e - y}}

Since the base is equal, we can equate the powers. On equating the powers, we have,
⇒y=e−y\Rightarrow y = e - y
Taking yy from R.H.S. to L.H.S.
⇒2y=e\Rightarrow 2y = e
Dividing both the sides by 22, we get
⇒y=e2−−−−(6)\Rightarrow y = \dfrac{e}{2} - - - - (6)
Now we will evaluate equation (5)\left( 5 \right) at x=ex = e and use equation (6)\left( 6 \right).
⇒(dydx)x=e=1−(e2)e1+ln⁡e\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = e}} = \dfrac{{1 - \dfrac{{\left( {\dfrac{e}{2}} \right)}}{e}}}{{1 + \ln e}}
Putting ln⁡e=1\ln e = 1 we get;
⇒(dydx)x=e=1−(e2)e1+1\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = e}} = \dfrac{{1 - \dfrac{{\left( {\dfrac{e}{2}} \right)}}{e}}}{{1 + 1}}
We can write (dydx)x=e{\left( {\dfrac{{dy}}{{dx}}} \right)_{x = e}} as y1(e){y^1}\left( e \right), therefore on simplification, we get
⇒y′(e)=1−122\Rightarrow y'\left( e \right) = \dfrac{{1 - \dfrac{1}{2}}}{2}
⇒y′(e)=14\Rightarrow y'\left( e \right) = \dfrac{1}{4}

So, the second option of list 1 matches with the second option of list 2.

Now we will take the third question.
x+y=ex−yx + y = {e^{x - y}}
Differentiating with respect to xx,
⇒ddx(x+y)=ddx(ex−y)\Rightarrow \dfrac{d}{{dx}}\left( {x + y} \right) = \dfrac{d}{{dx}}\left( {{e^{x - y}}} \right)
Using the chain rule of differentiation in R.H.S., we get
⇒1+dydx=ex−yd(x−y)dx\Rightarrow 1 + \dfrac{{dy}}{{dx}} = {e^{x - y}}\dfrac{{d\left( {x - y} \right)}}{{dx}}
⇒1+dydx=ex−y(1−dydx)\Rightarrow 1 + \dfrac{{dy}}{{dx}} = {e^{x - y}}\left( {1 - \dfrac{{dy}}{{dx}}} \right)
Now we will simplify the R.H.S.
⇒1+dydx=ex−y−ex−ydydx\Rightarrow 1 + \dfrac{{dy}}{{dx}} = {e^{x - y}} - {e^{x - y}}\dfrac{{dy}}{{dx}}
Taking ex−ydydx{e^{x - y}}\dfrac{{dy}}{{dx}} from R.H.S to L.H.S. and 11 from L.H.S to R.H.S., we get
⇒dydx+ex−ydydx=ex−y−1\Rightarrow \dfrac{{dy}}{{dx}} + {e^{x - y}}\dfrac{{dy}}{{dx}} = {e^{x - y}} - 1
Taking dydx\dfrac{{dy}}{{dx}} common from L.H.S., we get
⇒(1+ex−y)dydx=ex−y−1\Rightarrow \left( {1 + {e^{x - y}}} \right)\dfrac{{dy}}{{dx}} = {e^{x - y}} - 1
Dividing both the sides by (1+ex−y)\left( {1 + {e^{x - y}}} \right), we get
⇒dydx=ex−y−1ex−y+1−−−−(7)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^{x - y}} - 1}}{{{e^{x - y}} + 1}} - - - - \left( 7 \right)

Differentiating (7)(7) with respect to xx,
⇒ddx(dydx)=ddx(ex−y−1ex−y+1)\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{e^{x - y}} - 1}}{{{e^{x - y}} + 1}}} \right)
Using; ddx(uv)=vdudx−udvdxv2\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}
Therefore, we get;
⇒d2ydx2=(ex−y+1)d(ex−y−1)dx−(ex−y−1)ddx(ex−y+1)(ex−y+1)2\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {{e^{x - y}} + 1} \right)\dfrac{{d\left( {{e^{x - y}} - 1} \right)}}{{dx}} - \left( {{e^{x - y}} - 1} \right)\dfrac{d}{{dx}}\left( {{e^{x - y}} + 1} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}}
On differentiating we have;
⇒d2ydx2=(ex−y+1)ex−yd(x−y)dx−(ex−y−1)ex−yd(x−y)dx(ex−y+1)2\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {{e^{x - y}} + 1} \right){e^{x - y}}\dfrac{{d\left( {x - y} \right)}}{{dx}} - \left( {{e^{x - y}} - 1} \right){e^{x - y}}\dfrac{{d\left( {x - y} \right)}}{{dx}}}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}}
On further differentiating we get;
⇒d2ydx2=(ex−y+1)ex−y(1−dydx)−(ex−y−1)ex−y(1−dydx)(ex−y+1)2\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {{e^{x - y}} + 1} \right){e^{x - y}}\left( {1 - \dfrac{{dy}}{{dx}}} \right) - \left( {{e^{x - y}} - 1} \right){e^{x - y}}\left( {1 - \dfrac{{dy}}{{dx}}} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}}

Taking ex−y(1−dydx){e^{x - y}}\left( {1 - \dfrac{{dy}}{{dx}}} \right) common from the numerator,
⇒d2ydx2=ex−y(1−dydx)(ex−y+1−ex−y+1)(ex−y+1)2\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{{e^{x - y}}\left( {1 - \dfrac{{dy}}{{dx}}} \right)\left( {{e^{x - y}} + 1 - {e^{x - y}} + 1} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}}
On solving the numerator, we have;
⇒d2ydx2=2ex−y(1−dydx)(ex−y+1)2−−−−(8)\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{2{e^{x - y}}\left( {1 - \dfrac{{dy}}{{dx}}} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} - - - - (8)
Now we have;
x+y=ex−yx + y = {e^{x - y}}
At x=0x = 0 we have,
⇒y(0)=e−y(0)−−−−(9)\Rightarrow y\left( 0 \right) = {e^{ - y\left( 0 \right)}} - - - - (9)
We will evaluate equation (7)(7) at x=0x = 0.
⇒(dydx)x=0=e−y(0)−1e−y(0)+1\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = \dfrac{{{e^{ - y\left( 0 \right)}} - 1}}{{{e^{ - y\left( 0 \right)}} + 1}}
Using equation (9)(9) we get;
⇒(dydx)x=0=y(0)−1y(0)+1−−−−(10)\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = \dfrac{{y\left( 0 \right) - 1}}{{y\left( 0 \right) + 1}} - - - - (10)

We will find the value of equation (8)(8) at x=0x = 0.
⇒(d2ydx2)x=0=2e−y(0)(1−(dydx)x=0)(e−y(0)+1)2\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{2{e^{ - y\left( 0 \right)}}\left( {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}_{x = 0}}} \right)}}{{{{\left( {{e^{ - y\left( 0 \right)}} + 1} \right)}^2}}}
Now we will use equation (9)(9),
\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{2y\left( 0 \right)\left\\{ {1 - \dfrac{{y\left( 0 \right) - 1}}{{y\left( 0 \right) + 1}}} \right\\}}}{{{{\left( {{e^{ - y\left( 0 \right)}} + 1} \right)}^2}}}
On simplification we get,
\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{2y\left( 0 \right)\left\\{ {\dfrac{{y\left( 0 \right) + 1 - y\left( 0 \right) + 1}}{{y\left( 0 \right) + 1}}} \right\\}}}{{{{\left( {y\left( 0 \right) + 1} \right)}^2}}}
\Rightarrow y''(0) = \dfrac{{2y\left( 0 \right)\left\\{ 2 \right\\}}}{{{{\left( {y\left( 0 \right) + 1} \right)}^3}}}
On rearranging we get,
⇒y′′(0)=4y(0)(1+y(0))3\Rightarrow y''(0) = \dfrac{{4y\left( 0 \right)}}{{{{\left( {1 + y\left( 0 \right)} \right)}^3}}}

Hence, the third question from list 1 matches with the fourth option in list 2.

The fourth question we have is;
ex−ey=y−x{e^x} - {e^y} = y - x
Differentiating it w.r.t xx, we get,
⇒ddx(ex−ey)=ddx(y−x)\Rightarrow \dfrac{d}{{dx}}\left( {{e^x} - {e^y}} \right) = \dfrac{d}{{dx}}\left( {y - x} \right)
On solving,
⇒ex−eydydx=dydx−1\Rightarrow {e^x} - {e^y}\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dx}} - 1
Differentiating it again we get;
⇒ddx(ex−eydydx)=ddx(dydx−1)\Rightarrow \dfrac{d}{{dx}}\left( {{e^x} - {e^y}\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}} - 1} \right)
⇒ex−ey(dydx)2−ey(d2ydx2)=d2ydx2\Rightarrow {e^x} - {e^y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {e^y}\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) = \dfrac{{{d^2}y}}{{d{x^2}}}
On taking ey(d2ydx2){e^y}\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) from L.H.S. to R.H.S.
⇒ex−ey(dydx)2=d2ydx2+ey(d2ydx2)\Rightarrow {e^x} - {e^y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \dfrac{{{d^2}y}}{{d{x^2}}} + {e^y}\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)
Taking d2ydx2\dfrac{{{d^2}y}}{{d{x^2}}} common from R.H.S. and on rewriting, we get
⇒(1+ey)d2ydx2=ex−ey(dydx)2\Rightarrow \left( {1 + {e^y}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = {e^x} - {e^y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2}
Dividing both the sides by (1+ey)\left( {1 + {e^y}} \right), we get
⇒d2ydx2=ex−ey(dydx)2(1+ey)−−−−(11)\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{{e^x} - {e^y}{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}}{{\left( {1 + {e^y}} \right)}} - - - - (11)

Now we have;
ex−ey=y−x{e^x} - {e^y} = y - x
Putting x=0x = 0 we get;
⇒1−ey(0)=y(0)−−−−(12)\Rightarrow 1 - {e^{y\left( 0 \right)}} = y\left( 0 \right) - - - - (12)
Also, we have from above:
⇒ex−eydydx=dydx−1\Rightarrow {e^x} - {e^y}\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dx}} - 1
On taking eydydx{e^y}\dfrac{{dy}}{{dx}} from L.H.S. to R.H.S. and 11 from R.H.S. to L.H.S.,
⇒ex+1=dydx+eydydx\Rightarrow {e^x} + 1 = \dfrac{{dy}}{{dx}} + {e^y}\dfrac{{dy}}{{dx}}
Taking dydx\dfrac{{dy}}{{dx}} common from R.H.S.
⇒ex+1=(1+ey)dydx\Rightarrow {e^x} + 1 = \left( {1 + {e^y}} \right)\dfrac{{dy}}{{dx}}
Dividing both the sides by (1+ey)\left( {1 + {e^y}} \right) and on rearranging we get
⇒dydx=ex+1ey+1\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^x} + 1}}{{{e^y} + 1}}
Finding its value at x=0x = 0 we get;
⇒(dydx)x=0=2ey(0)+1−−−−(13)\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = \dfrac{2}{{{e^y}^{\left( 0 \right)} + 1}} - - - - (13)

Now finding the value of equation (11)(11) at x=0x = 0.
⇒(d2ydx2)x=0=1−ey(0)(dydx)x=021+ey(0)\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{1 - {e^{y\left( 0 \right)}}\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}^2}}{{1 + {e^{y\left( 0 \right)}}}}
Now using equation (12),(13)\left( {12} \right),\left( {13} \right) we get;
⇒y′′(0)=1−y(0)(2ey(0)+1)21+ey(0)\Rightarrow y''\left( 0 \right) = \dfrac{{1 - y\left( 0 \right){{\left( {\dfrac{2}{{{e^{y\left( 0 \right)}} + 1}}} \right)}^2}}}{{1 + {e^{y\left( 0 \right)}}}}
On evaluating further, we get;
⇒y′′(0)=1−4ey(0)(ey(0)+1)21+ey(0)\Rightarrow y''\left( 0 \right) = \dfrac{{1 - \dfrac{{4{e^{y\left( 0 \right)}}}}{{{{\left( {{e^{y\left( 0 \right)}} + 1} \right)}^2}}}}}{{1 + {e^{y\left( 0 \right)}}}}
On solving the numerator further, we will get;
⇒y′′(0)=e2y(0)+1+2ey(0)−4ey(0)(1+ey(0))3\Rightarrow y''\left( 0 \right) = \dfrac{{{e^{2y\left( 0 \right)}} + 1 + 2{e^{y\left( 0 \right)}} - 4{e^{y\left( 0 \right)}}}}{{{{\left( {1 + {e^{y\left( 0 \right)}}} \right)}^3}}}
⇒y′′(0)=e2y(0)+1−2ey(0)(1+ey(0))3\Rightarrow y''\left( 0 \right) = \dfrac{{{e^{2y\left( 0 \right)}} + 1 - 2{e^{y\left( 0 \right)}}}}{{{{\left( {1 + {e^{y\left( 0 \right)}}} \right)}^3}}}
We can see that the numerator is in the form of (a−b)2{\left( {a - b} \right)^2}. So, solving further we get;
⇒y′′(0)=(1−ey(0))2(1+ey(0))3\Rightarrow y''\left( 0 \right) = \dfrac{{{{\left( {1 - {e^{y\left( 0 \right)}}} \right)}^2}}}{{{{\left( {1 + {e^{y\left( 0 \right)}}} \right)}^3}}}

Therefore, the fourth question from list 1 matches with the first option.

Note: One important point to note is that while solving the questions we need to keep in mind the options because we have to get the results as given in the options. As you can see that in the last step of the fourth question, we might write e2y(0)+1−2ey(0)=(ey(0)−1)2{e^{2y\left( 0 \right)}} + 1 - 2{e^{y\left( 0 \right)}} = {\left( {{e^{y\left( 0 \right)}} - 1} \right)^2}, but that’s not given in the option. So, in the match the column types question we have to solve the question keeping in mind the given options.