y = log(x²+1)+c

\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)

Equation of family of straight lines in my plane is y = mx - c where m and c are arbitrary constraints.
Differentiating, y' = m
Differentiating again, y" = 0, is the required differential equation.

y = Acos x + Bsin x ... (1)
Differentiating (1) twice successively, we get
\(\frac{dy}{dx}\)= −Asin x + Bcos x. ...(2)
\(\frac{d^2y}{dx^2}\) = -Acos x − Bsin x = −(A cos x + B sin x). ...(3)
Substituting (1) in (3), we get \(\frac{d^2y}{dx^2}\) + = 0 as the required differential equation

The given differential equation is \(3\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 3 }{ 2 } }\)Squaring both sides, we get 
\(9{ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 3 }\)
In this equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 2.
Therefore, the given differential equation is of order 2 and degree 2.

y''− 5y' + 6y = 0 ......(1)
Given y = e^mx  .....(2)
Differentiating cquation (2) w.r.t 'x', we get
\(\frac{dy}{dx} = em^x . m\)
To find the value of m:
Given y" - 5y' + 6y = 0 
e^mx . m² -5e^mx+ 6e^mx = 0
e^mx [m- 5m +6] = 0
m - 5m + 6 = 0
(m - 3) (m - 2) = 0
m = 3, 2

Given = e^mx is the solution of
y' + 2y = 0 ...(1)
y = e^mx  ...... (2) 
\(\frac{dy}{dx} = e^{mx}. m\)
\(\frac{dy}{dx} = ym\)
\(\frac{dy}{dx} - my=0\)
⇒ y' - my = 0 ...(3)
Comparing equation (1) & (3),
we get m = -2

\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=0\)
The given differential equation is
\({ x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \) = -\({ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }\)
= - \(\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides,
\({ x }^{ 4 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) =1+{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
The highest derivative is 2 and its power is 2.
∴ Order 2, degree 2.

\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +4=0\)
Given differential equation is
\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) +5\frac { dy }{ dx } +4=0\)
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }=3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4\)
Taking power 3 both sides,
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }={ \left( 3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4 \right) }^{ 3 }\)
The highest derivative is 3 and its power is 2.
∴ Order 3, degree 2.

Let x(t) denote the amount of salt in the tank at time t.
Its rate of change is
\(\frac{dx}{dt}\) = inflow rate - outflow rate
Now, 2 gram time 3 litres per minutes is inflow rate = 6 grams of salt. (3 x 2 = 6)
The out flow of salt is \(\frac{3}{50}\) times x = \(\frac{3x}{50}\)
\(\therefore \frac { dx }{ dt } =6-\frac { 3x }{ 50 } =\frac { 300-3x }{ 50 } \)
\(=-\frac { 3(x-100) }{ 50 } \)
\(\Rightarrow \frac { dx }{ x-100 } =-\frac { 3 }{ 50 } dt\)
\(\Rightarrow \int { \frac { dx }{ x-100 } =-\frac { 3 }{ 50 } \int { dt } } \)
\(\Rightarrow log(x-100)=-\frac { 3 }{ 50 } t+logC\)
\(\\ \Rightarrow log(x-100)-logC=-\frac { 3 }{ 50 } t\)
\(\Rightarrow log\left( \frac { x-100 }{ C } \right) =-\frac { 3 }{ 50 } t\)
\(\Rightarrow \frac { x-100 }{ C } ={ e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow x-100={ C }_{ e }-\frac { 3t }{ 50 } \quad ...(1)\)
When t = 0, x = 0
[Since initial water was pure without any salt]
\(\Rightarrow\) 0-100 = Ce⁰ 
\(\Rightarrow\) C = -100
(1) becomes x-100 = -100\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow\) x = 100-100\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow\) x = 100(1-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))
Hence the amount of salt in the tank at time t is x = 100(1-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))

Let there be N radioactive nuclei in a sample at any time t and let N_o be the initial number of radioactive nuclei.
Then \(\frac{dN}{dt}\infty N\)
\(\Rightarrow \frac { dN }{ dt } =-\lambda N\)
Where \(\lambda>0\) is a constant
\(\Rightarrow \frac { dN }{ N } =-\lambda dt\)
\(\int { \frac { dN }{ N } } =-\lambda dt\)
\(\int { \frac { dN }{ N } =-\lambda \int { dt } } \)
\(\Rightarrow log\ N=-\lambda t+C\ ...(1)\)
\(T\quad t=0,\ we\ have\ N={ N }_{ 0 }\)
\(\therefore log{ N }_{ 0 }=0+C\)
\(\Rightarrow C=log{ N }_{ 0 }\)
\(\therefore\)(1) becomes, log N = -\(\lambda t+log{ N }_{ 0 }\)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } =-\lambda t\quad ...(2)\)
It is given that 10% of the original number of nuclei have undergone disintegration in a period of 100 years.
Whent= 100 \(N={ N }_{ 0 }-\frac { 10 }{ 100 } \times { N }_{ 0 }=\frac { { 9N }_{ 0 } }{ 10 } \)
Substituting in (2) we get
\(log\quad \frac { 9 }{ 10 } =-100\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 100 } log\frac { 9 }{ 10 } \)
Substituting in (2) we get,
\(log\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 100 } log\frac { 9 }{ 10 } \right) t\)
when t = 1000,
\(log\frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 100 } log\left( \frac { 9 }{ 10 } \right) \times 1000\)
\(=10log\left( \frac { 9 }{ 10 } \right) \)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } \times 100={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\times 100=\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \)
Hence, \(\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \%\) of radioactive nuclei will remain after 1000 years,

Let x (t) denote the amount of salt in the tank at time t. Its rate of change is \(\frac{dx}{dt}\) = in flow rate − out flow rate
Now, 5 grams times 10 litres gives an inflow of 50 grams of salt. Also, the out flow of brine is 10 litres per minute. This is 10 /1000 = 0.01of the total brine content in the tank. Hence, the out flow of salt is 0.01 times x(t) , that is 0.01x(t).
Thus the differential equation for the model is \(\frac{dx}{dt}\)= 50 − 0.01x = −0.01(x − 5000)
This can be written as \(\frac{dx}{x-5000}=-(0.01)dt\)
Integrating both sides, we obtain log |x − 5000| = −0.01t + log C
or x-5000 = Ce^-0.01t or x = 5000+Ce^-0.01t
Initially, whent = 0, x = 100 , so 100 = 5000 +C .Thus, C = −4900 .
Hence, the amount of the salt in the tank at time t is x = 5000 -4900e^-0.01

Let A be the mass of the isotope remaining after t years, and let −k be the constant of proportionality, where k > 0. Then the rate of decomposition is modeled by \(\frac{da}{dt}=-kA,\) where the minus sign indicates that the mass is decreasing. It is a separable equation. Separating the variables,we get\(\frac{da}{dt}=-kdt\).
Integrating on both sides, we get log |A| = −kt + log |C| or A = Ce−^kt.
Given that the initial mass is 200mg. That is, A = 200 when t = 0 and thus, C = 200.
Thus, we get A = − 200e^-kt.
Also, A =150when t = 2 and therefore, k = \(\frac{1}{2}log(\frac{4}{3})\)
Hence, A(t) = 200e^{\(\frac{1}{2}log(\frac{4}{3})\)} is the mass of isotope remaining after t years.
The half-life t_h is the time corresponding to A = 100 mg
Thus, \({ t }_{ k }=\frac { 2log\left( \frac { 1 }{ 2 } \right) }{ log\left( \frac { 3 }{ 4 } \right) } \).

\(\frac { dy }{ dx } +\frac { { 3x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation 
\(\therefore P=\frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
\(\therefore \int { pdx } =\int { \frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =log(1+{ x }^{ 3 })\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log(1+{ x }^{ 3 }) }=(1+{ x }^{ 3 })\)
\(\therefore\)The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } (1+{ x }^{ 3 })dx+c } \)
\(cos2x=1-2{ sin }^{ 2 }x\)
\(sin2x=\frac { 1-cos2x }{ 2 } =\int { { sin }^{ 2 }xdx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { 1-cos2x }{ 2 } } dx+c\)
\(\Rightarrow y(1+{ x }^{ 3 })=\frac { x }{ 2 } -\frac { sin2x }{ 4 } +c\)

\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } .cot\left( \frac { x }{ y } \right) =xdx\)
put \(\frac { x }{ y } =t\)
\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } =dt\)
Substituting these values in equation (1), we get
dt cot(t) = x dx
 cot t dt = ndx
Taking integration on both sides, we get
\(\Rightarrow \int { cot(t)dt=n\int { dx } } \)
\( \int \cot t \mathrm{dt} =n \int d x \)
\(\log (\sin \mathrm{t}) =\mathrm{n} x+\mathrm{C}_1 \)
\(\sin \mathrm{t} =\mathrm{e}^{\mathrm{nx}+\mathrm{c}_1} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{e}^{n x} \mathrm{e}^{\mathrm{C}_r} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{C}^{\mathrm{nx}}\)
\(\\ \Rightarrow sin\left( \frac { x }{ y } \right) ={ e }^{ nx+c }\left[ \because t=\frac { x }{ y } \right] \)

Given \(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
On separating the variables we get,
\(\frac { vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { g }{ { k }^{ 2 } } .dx\)
Multiplying by -2 both sides we get,
\(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
\(\frac { -2vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { -2g }{ { k }^{ 2 } } dx\)
Taking integrating on both sides, we get
\(\int { \frac { -2v }{ { k }^{ 2 }-{ v }^{ 2 } } } =\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })=\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })-log\quad c=\frac { -2g }{ { k }^{ 2 } } .x\)
\(\Rightarrow log\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ c } \right) -\frac { -2gx }{ { k }^{ 2 } } \)
\(\Rightarrow \frac { { k }^{ 2 }-{ v }^{ 2 } }{ e } ={ e }^{ -\frac { -2gx }{ { k }^{ 2 } } }\)
\(\Rightarrow { k }^{ 2 }-{ v }^{ 2 }{ ce }^{ \frac { -2gx }{ { k }^{ 2 } } }...(1)\)
Initial condition:
when v = 0, x = 0 we get
\( k^2-(0)^2 =C e^{\frac{-2g(0)}{k^2}} \)
\(k^2 =C e^0 \Rightarrow k^2=C \)
\((1) \Rightarrow k^2-v^2 =k^2 e^{\frac{-2 x^2}{k^2}} \)
\(k^2-k^2 e^{\frac{-2 s x}{k^1}} =\mathrm{v}^2 \)
\(k^2\left[1-e^{\frac{-2 s x}{k^2}}\right] =\mathrm{v}^2\)

Given equation is m \(\frac{dV}{dt}\) = F- kv  
The given equation can be written as
\(\frac { dv }{ F-kv } =\frac { dt }{ m } \)
Now Integrating, we get
\(\int { \frac { dv }{ F-kv } } =\int { \frac { dt }{ m } } \)
\(\int \frac{d V}{F-k V}=\int \frac{d t}{M}   \frac{\log (F-k V)}{-k}=\frac{t}{M}+C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}-k C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}+\log \mathrm{C} \\ \log [\mathrm{F}-\mathrm{kV}]-\log \mathrm{C}=-\frac{k t}{M} \\ \log \left(\frac{F-k V}{C}\right)=-\frac{k t}{M} \\ \frac{F-k V}{C}=e^{\frac{-k t}{M}} \\ \frac{F-k V}{e^{\frac{-t}{M}}}=\mathrm{C} \Rightarrow \mathrm{C}=e^{\frac{k t}{M}}(F-k V) \\\)
Initial condition:
Given V = 0 when t = 0
\(\mathrm{C}=e^{\frac{k(0)}{M}}[\mathrm{~F}-\mathrm{k}(0)] \\ =\mathrm{e}^0[\mathrm{~F}-0] \\ \mathrm{C}=\mathrm{F} \\ \therefore \mathrm{F}=(F-k V) e^{\frac{k t}{M}}\)

\(xy=\frac { { y }^{ 4 } }{ 4 } +c\)

prove

\(\left( \frac { dy }{ dx } \right) ^{ 2 }=4y\left( x\frac { dy }{ dx } -2y \right) \)

Given \(\frac { dy }{ dx } +\frac { { y }^{ 2 } }{ { x }^{ 2 } } =\frac { y }{ x } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y }{ x } -\frac { { y }^{ 2 } }{ { x }^{ 2 } } ...(1)\)
This is a homogeneous differential equation
put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) becomes,
\(v+x\frac { dv }{ dx } =\frac { vx }{ x } -\frac { { v }^{ 2 }{ x }^{ 2 } }{ { x }^{ 2 } } =v-{ v }^{ 2 }\)
\(x\frac { dv }{ dx } =v-{ v }^{ 2 }-v=-v\)
\(\therefore \frac { dv }{ { v }^{ 2 } } =\frac { -dx }{ x } \)
\(\Rightarrow \int { \frac { dv }{ { v }^{ 2 } } =-\int { \frac { dx }{ x } } } \)
\(\Rightarrow \frac { -1 }{ v } =-log\quad x+c\)
\(\Rightarrow \frac { -1 }{ \frac { y }{ x } } =-log\quad x+c\)
\(\Rightarrow \frac { -x }{ y } +logx=c\)

The given equation is rewritten as \(\frac { dy }{ dx } =\frac { { y }^{ 2 } }{ xy-{ x }^{ 2 } } \)
This is a homogeneous differential equation 
Put y = vx . Then, we have \(x\frac { dv }{ dx } =\frac { v }{ v-1 } \)
By separating the variables, \(\frac { v-1 }{ v } dv=\frac { dx }{ x } .\)
Integrating, we obtain v − log |v| = log |x| + log |C| or v = log |vxC|.
Replacing v by \(\frac{y}{x}\), we get, \(\frac{y}{x}\) = log |Cy| = e^y/x or y = ke^y/x (how!) which is the required solution.

\(sin\left( \frac { dy }{ dx } \right) =a\)
\(\Rightarrow \frac { dy }{ dx } ={ sin }^{ -1 }(a)\)
\(\Rightarrow dy={ sin }^{ -1 }(a)dx\)
\(\Rightarrow \int { dy } ={ sin }^{ -1 }(a)\int { dx } \)
Taking Integration on both sides, we get
\(\Rightarrow \int { dy } ={ sin }^{ -1 }(a)\int { dx } \)
\(\Rightarrow y={ sin }^{ -1 }(a)x+c ...(1)\)
Initial condition:
Since y(0) = 1 we get,
1 = sin^-1(a)(0) +C
0 + C ⇒ C = 1
equation (1) ⇒ y = sin^-1(a) x + 1
y^-1 = sin^-1(a) + x
\(\Rightarrow \frac { y-1 }{ x } =sin(a)\Rightarrow sin\left( \frac { y-1 }{ x } \right) =a\)

given equation of family of curves is
y = Ae^8x + Be^-8x ..(1)
where A & B are arbitrary constants. Differentiating cquation (1) twice successively (because we have two arbitrary constant), we get
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } \\ \\ \) = 8Ae^8x + 8Be^-8x
Differentiating again with respect to 'x' we get,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64Ae^8x + 64Be^-8x
= 64(Ae^8x + Be^-8x)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64 y [using (1)]
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) - 64 y = 0
Which is the required differential equation.

The equation of the family of ellipses having centre at the origin & foci on the y-axis, is given
\(\frac { { x }^{ 2 } }{ { b }^{ 2 } } +\frac { { y }^{ 2 } }{ { a }^{ 2 } } =1\) ...(1)
where b >a & a, b are the parameters or a,b are arbitrary constant.
Differentiating equation (1) twice successively, because we have two arbitrary constant) we get
\( \frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \frac{d y}{d x} =0 \)
\(2\left(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}\right) =0 \)
\(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x} =0\) ............(2)
Again differentiating equation (2)
\(\frac{1}{a^{2}}+\frac{y}{b^{2}} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \frac{d y}{d x b^{2}}=0\)
\(\frac{1}{a^{2}}+\frac{y}{b^{2}} \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2} \frac{1}{b^{2}}=0\)
multiply by x
\(\frac{x}{a^{2}}+\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2} \frac{x}{b^{2}}=0\) .........(3) 
Equation (3)-(2)
\( \frac{x}{a^{2}}+\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2}\left(\frac{x}{b^{2}}\right) -\left(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}\right) =0 \)
\(\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2} \frac{x}{b^{2}}-\frac{y}{b^{2}} \frac{d y}{d x} =0 \)
Taking \(\frac{1}{b^{2}}\) outside, we get
\( \frac{1}{b^{2}}\left[x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}\right]=0 \\ x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0 \)
is the required differential equation.

The equation of the family of  parabolas with latus rectum (4a) and whose axes are parallel to the x-axis is shown in sketch
Let vertex 'V' be (h, k) and focus at 'F" and let
L-L' be latus rectum = 4a
Hence, FL = 2a and VF = a.
Thus 'F' is at (h+a, k).

Equation of parabola with vertex at (h, k) and focal length 'a', latus rectum 4a' is
(y - k)² = 4a(x - h) .......(1)
Differentiating with respect to 'x' we got
2(y - k) \(\frac{dy}{dx}\) = 4a 
⇒ (y - k). \(\frac{dy}{dx}\) = 2a ....... (2)
Again differentiating with respect to x,
(y-k) y''+y'\(\times\) y' = 0  ............(3)
From(2),(y-k) = From(2),(y-k) = \(\frac{2a}{y'}\)
Putting in (3), we get
\((\frac{2a}{y'})y''+(y')^2=0 (or) 2ay'' +(y')^3 = 0\)
This is the required differential equation.