The given function is y mx +\(\frac{7}{m}\), where m is an arbitrary constant ....(1)
Differentiating both sides of equation (1) with respect to x, we get y' = m.
Substituting the values of y' and y in the given differential equation
we get xy'\(\frac{1}{y'}\)-y = xm +\(\frac{7}{m}\)- mx -\(\frac{7}{m}\) = 0
Therefore, the given function is a solution of the differential equation xy' + 7\(\frac{1}{y'}\) - y = 0

Given that x² + y² = r², r∈R      ...(1)
The given equation contains exactly one arbitrary constant.
So, we have to differentiate the given equation once. Differentiate (1) with respect to x, we get
2x +2y\(\frac{dy}{dx}\) = 0  which implies \(\frac{dy}{dx}\) = \(-\frac{x}{y}\)
Thus, x² + y² = r² satisfies the differential equation \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).
Hence, x² + y² = r² is a solution of the differential equation  \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).

The equation of the family of parabolas is given by y² ax = 4, a is an arbitrary constant. ... (1)
Differentiating both sides of (1) with respect to x , we get 2y\(\frac{dy}{dx}=4a\Rightarrow a=\frac{y}{2}\frac{dy}{dx}\) 
Substituting the value of a in (1) and simplifying, we get \(\frac{dy}{dx}=\frac{y}{2x}\) as 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 family of straight lines passing through the origin is y = mx, where m is an arbitrary constant.
Differentiating both sides with respect to x, we get \(\frac{dy}{dx}=m\) 
From (1) and (2), we get y = x\(\frac{dy}{dx}\). This is the required differential equation. 
Observe that the given equation y = mx contains only one arbitrary constant and thus we get the differential equation of order one.

dy + (xy − cos x)dx = 0 is a first order differential equation with degree 1 
since the equation can be rewritten as 
\(\frac{dy}{dx}\) + xy - cos x = 0

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.

In the given differential equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 1.
Therefore, the given differential equation is of order 2.
The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.

Here, the highest order derivative is \(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } \) whose power is 3.
Therefore, the given differential equation is of order 4 and degree 3.

In this equation, the highest order derivative is \(\\ \\ \\ \frac { dy }{ dx } \) whose power is 1.
Therefore, the given differential equation is of order 1 and degree 1.

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

y = ae^x + be^−x; y − y = 0
Consider y = ae^x + be^−x  
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } = ae^{x} -be^{-x} \)
Again differentiating, we get
\(\frac { d^2y }{ dx^2 } = ae^{x} -be^{-x} \)
\(\frac { d^2y }{ dx^2 } = y\)
\(\frac { d^2y }{ dx^2 } - y = 0\) is a required differential equation.
Thus y = ae^x + be^−x satisfies the diferential| equation y"- y = 0
Hence y = ae^x + be^−x is a solution of the differential equation y"- y = 0

y = 2x² ; xy' = 2y
Consider y = ax² ...(1)
Differentiating with respect to 'x' we get,
⇒ \(\frac { dy }{ dx } = 2.2x\)  ...(2)
Multiply by x on both sides, we get,
x = \(\frac { dy }{ dx } = 2.2x\) 
y' = 2y is a given differential equation
Thus y = 2x² satislies the differential equation
xy' = 2y 
Hence y = 2x² is a solution of the differential equation xy' = 2y.

Given equation of curve is
xy = ae^x + be^−x + x² ...(1)
where a &b are aribitrary constant. differentiate equation (1) twice successively, because we have two arbitray constant.
\(x \frac{d y}{d x}+y(1)=a \mathrm{e}^{x}-\mathrm{be}^{-x}+2 x\) ...(2)
\( x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(1)+\frac{d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \)
\( x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \) ...(3)
From (1), we get \(x y-x^{2}=\mathrm{ae}^{x}+\mathrm{be}^{-x}\) ...(4)
Substituting equation (a) in (3), we get
\(\therefore x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}-x y+x^{2}-2=0\) which is the required differential equaiton. 

Let x represent the principal in the saving amount.
R = 8% and N = 1.
∴ Interest = \(\frac { PNR }{ 100 } =\frac { x\times 1\times 8 }{ 100 } =\frac { 2x }{ 25 } \)
∴ Given \(\frac { dx }{ dt } \) = interest + Rs. 400.
∴ \(\frac { dx }{ dt } =\frac { 2x }{ 25 } +400\)

Let P represent the vapour pressure and T represent the vapour temperature.
Given \(\frac { dp }{ dt } \alpha \quad p.\frac { 1 }{ { T }^{ 2 } } \)
[∴ Inversely proportional to the square of the  temperature]
\(\Rightarrow \frac { dp }{ dt } =\frac { kP }{ { T }^{ 2 } } \) where k is a constant.

(i) If at any time t, The amount of Radium present is Q. The rate at which Q is decreasing \(\frac { dQ }{ dt } \).
This rate of decrease or decay is found to be proportional to Q itself. Hence we have the law, \(\frac { dQ }{ dt } = kQ\). where k is the dt constant of proportionality. Which is a required differential equation.
(ii) The rate of change of population Solution increases with respect to time t, is \(\frac { dp }{ dt } \) & the rate of population is proportional| the product of population is \(\frac { dp }{ dt } \) = kP & the also the difference between 5,00,000 & the population is \(\frac { dp }{ dt } \) = kP (5,00,000 - P) is a  required differential equation.
(iii) The rate of change of vapor pressure P with respect to time t is \(\frac { dp }{ dt } \)& the rate of dt increase vapor pressure is P at time T is proportional to the vapor pressure and also is inversely proportional to the square of the temperature is \(\frac { dp }{ dt } \)\(\infty\) P and \(\frac { dp }{ dt } ​​\infty\frac{1}{T^2}\)
Combining the two, we get
\(\frac { dp }{ dt } ​​\infty\frac{p}{T^2} \Rightarrow \frac { dp }{ dt }= k(\frac{p}{T^2})\), where 'k' is constantof proportionality
(iv) Let x be the amount. Amount varies from every year. (ie) Amount varies with respect to time t is \(\frac { dp }{ dt } \) & in addition the income from other source credited Rs. 400 continuously for every year.
\(\frac { dx }{ dt } = \frac{8}{100}\times x + 400\)
\(\Rightarrow\frac{dx}{dt} = \frac{2x}{25}+400\) is a required differential equation.

\(x={ e }^{ xy\left( \frac { dy }{ dx } \right) }\)
Taking log on both sides, log x = xy \(\frac{dy}{dx}\)
In this equation, the highest-order derivative is \(\frac{dy}{dx}\) so its power is 1.
The highest derivative is 1.
∴ its Order = 1, Degree = 1

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +\int { ydx } ={ x }^{ 3 }\)
The given differential equation is
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +\int { ydx } ={ x }^{ 3 }.\)
Differentiating again with respect to 'x' we get.
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y=3{ x }^{ 2 }\)
The highest derivative is 3 and its power is 1.
∴ Order is 3 and degree is 1.

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =xy+cos\left( \frac { dy }{ dx } \right) \)
The given differential equation is
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =xy+cos\left( \frac { dy }{ dx } \right) \)
The highest derivative is 2
∴ Order 2.
The given differential equation is not a polynomial equations in its derivatives and so its degree is not defined.

The given differential equation is 
\({ \left( \frac { d^2y }{ dx^2 } \right) }^{ 3\times2 }= { 1+\left( \frac { dy }{ dx } \right) } \)
squaring both sides, we get
\({ \left( \frac { dy }{ dx } \right) }^{ 6 }=1+\left( \frac { dy }{ dx } \right) \)
In this equation, the highest order derivative is 2 and its power is 6.
∴ Order 2, degree 6.

\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=0\)
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.

\(y\left( \frac { dy }{ dx } \right) =\frac { x }{ \left( \frac { dy }{ dx } \right) +{ \left( \frac { dy }{ dx } \right) }^{ 3 } } \)
is the given differential equation.
\(\Rightarrow y{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 4 }=x\)
The highest derivative is 1 and its maximum power is 4.
∴ Order 1, degree 4.

\(\sqrt { \frac { dy }{ dx } } -4\frac { dy }{ dx } -7x=0\)
The given differential equation is 
\(\sqrt { \frac { dy }{ dx } } =4\frac { dy }{ dx } +7x\)
Squaring both sides,
\(\frac { dy }{ dx } =\quad { \left( 4\frac { dy }{ dx } +7x \right) }^{ 2 }\)
\(16{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ 49 }x^{ 2 }+56x{ \left( \frac { dy }{ dx } \right) }\)
The highest derivative is 1 and its maximum power is 2.
∴ Order 1, degree 2.

\({ { \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) } }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 2 }=xsin\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \)
The highest derivative is 2
∴ Order 2
The given differential equation is not a polynomial equation in its derivative and so its degree is not defined.

\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +4=0\)
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.

\(\frac { dy }{ dx } +xy=cotx\)
Given differential equation is
\(\frac { dy }{ dx } +xy=cotx\)
The highest derivative is 1 and its power is 1 order 1, degree 1.

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.

To solve the given differential equation, we make the substitution 3x + y + 4 = z.
Differentiating with respect to x, we get \(\frac { dy }{ dx } =\frac { dz }{ dx } \)-3. 
So the given differential equation becomes \(\frac { dz }{ dx } \) = z²+ 3.
In this equation variables are separable. So, separating the variables and integrating, we get the general solution of the given differential equation as \(\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\left( \frac { 3x+y+4 }{ \sqrt { 3 } } \right) =x+C\)

The given function is y = a cos(log x) + bsin (log x)   ...(1)
where a, b are two arbitrary constants. In order to eliminate the two arbitrary constants, we have to differentiate the given function two times successively
Differentiating equation (1) with respect to x , we get 
y' = -a sin (log x).\(\frac{1}{x}\)+ b cos (log x).\(\frac{1}{x}\Rightarrow\)xy' = -a sin(log x)+b cos (log x).
Again differentiating this with respect to x, we get
xy" + y' = -a cos(log x).\(\frac{1}{x}-b\) sin (log x).\(\frac{1}{x}\Rightarrow\)x²y" + xy'+ y = 0
Therefore, y = a cos(log x) + bsin (log x) is a solution of the given differential equation.

The given function is y = 2(x²−1) + \(Ce^{x^2}\), where C is an arbitrary constant    ... (1)
Differentiating both sides of equation (1) with respect to x, we get \(\frac { dy }{ dx } =4x-2x{ Ce }^{ -x2 }\)
Substituting the values of \(\frac { dy }{ dx } \) and y in the given differential equation, we get
\(\frac { dy }{ dx } \) + 2xy - 4x³ = 4x - 2xCe^-x2 + 2x[2(x²-1)+Ce^-x2]-4x³ = 0
Therefore, the given function is a solution of the differential equation \(\frac { dy }{ dx } \)+2xy-4x³ = 0

The equation of the family of all ellipses having foci on the x -axis and centre at the origin is given by \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1,a>b\).
where a and b are arbitrary constants.
Differentiating equation (1) with respect to x, we get
\(\frac { 2x }{ { a }^{ 2 } } +\frac { 2y }{ { b }^{ 2 } } \frac { dy }{ dx } =0\quad \Rightarrow \frac { x }{ { a }^{ 2 } } +\frac { y }{ { b }^{ 2 } } \frac { dy }{ dx } =0\)
Differentiating equation (2) with respect to x, we get 
\(\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] =0\Rightarrow \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] \)
Substituting the value of \(\frac{1}{a^2}\) in equation (2) and simplifying, we get \(-\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] x+\frac { v }{ { b }^{ 2 } } \frac { dy }{ dx } =0\Rightarrow xy\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 }-y\frac { dy }{ dx } =0\) which is the required differential equation

The slope of the tangent to the curve at any point = \(\frac { 1 }{ 4(odinate\ at\ the\ point) } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { 1 }{ 4y } \)
The equation can be written as
\(\Rightarrow 4ydy\ =\ dx\) ....(1)
Integrating equation (1) on both sides, we get
\(4\int { y\quad dy } =\int { dx } \)
\(\Rightarrow 4.\frac { { y }^{ 2 } }{ 2 } =x+c\)
\(\Rightarrow { 2y }^{ 2 }=x+c\) ...(2)
Since the curve passes through (2, 5), we get
2(5)² = 2 + c
⇒ 50 - 2 = c
⇒ c = 48
Substituting the value of cincquation (2), we get
 2y² = x + 48 which is the requaired equation of the curve.

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.

Equation of family of parabolas with axis as y axis is given by,
(x-0) = 4a(y-k) .... (1)

Given: Vertex at (0, - 1).
Putting k = -1 in (1), we get
⇒ x² = \(\pm\)4a(y + 1) ....(2)
Differentiating with respect to 'x'  
2x = \(\pm\)4a\(\left( \frac { dy }{ dx } \right) \) ....(3)
⇒ 4a = \(\frac { 2x }{ \frac { dy }{ dx } } \) 
\(\frac{x^2}{2x} = \frac{y+1}{\frac{dt}{dx}}\)
ie) \(x \frac{dy}{dx}-2(y+1) =0\)
This 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.

Given the circles centre on r-axis & the circle is passing through the origin.
Let it be (r, 0) & its radius r.

Equation of the circle is
(x - a)² + (y - b)² = r²
(x - r)² + (y - 0)² = r²
⇒ x² - 2xr + r² + y² = r²
⇒ x² - 2xr + y² = 0 ...(1)
defferentiating equation (1) with respect to 'x' we get
⇒ 2x - 2r + 2y \(\frac { dy }{ dx } =0\)
⇒ 2x + 2y \(\frac { dy }{ dx } =2r\)
⇒ x + y \(\frac { dy }{ dx } =r\) ...(2)
Substituting r value in equation (1), we get
x² - 2x \(\left( x+y\frac { dy }{ dx } \right) +{ y }^{ 2 }=0\)
\(\Rightarrow \ { x }^{ 2 }-{ 2x }^{ 2 }-2xy\frac { dy }{ dx } +{ y }^{ 2 }=0\)
\(\Rightarrow \ { -x }^{ 2 }{ -2x }y\left( \frac { dy }{ dx } \right) { +y }^{ 2 }\)
Multiply by '-', we get
\(\Rightarrow \ { x }^{ 2 }{ +2x }y\left( \frac { dy }{ dx } \right) { -y }^{ 2 }\) which is the required differential equation.

Given circle equation be x² y² = r²
Let y = mx + c be the family of lines which touches the circle.
The condition for y = mx + c be all straight lines which towards the given circle x² y² = r² (1 + m²)
\(c=\sqrt { { 1+m }^{ 2 } } \)  
Hence, equation of tangent to the circle is .......(1)
y = mx + r\(\sqrt { { 1+m }^{ 2 } } \) .......(1) 
Differentiating with respect to x,
\(\frac { dy }{ dx } =m\quad ...(2)\)
Substituting 'm' in (1) we get,
\(y= \left( \frac { dy }{ dx } \right) \times x \pm r\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
\(y-x\left( \frac { dy }{ dx } \right) =\pm r\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides we get,
\(\Rightarrow { \left[ y-x\left( \frac { dy }{ dx } \right) \right] }^{ 2 }={ r }^{ 2 }\left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] \)
This is the differential equation of all straight lines touching the circle x² y² = r²

General equation of a straight line in a plane is ax + by = 1  .....(1)
Since, the lines are non - horizontal, a ≠ 0 
Hence, differentiating with respect to y, equation (1)
a \(\\ \frac { dx }{ dy }+b =0\)
Differentiating again with respect to 'y'; we get,
\((a) \ \frac { { d }^{ 2 }x }{ d{ y }^{ 2 } } =0\Rightarrow \frac { { d }^{ 2 }x }{ d{ y }^{ 2 } } =0\quad [\because a\neq 0]\)
This is the differential equation of all non-horizontal lines in a plane.

General equation of a straight line is 
ax + by + c = 0 .......(1)
where a, b, c \(\in\) R.
Since, the lines are non - vertical,we have b \(\neq\) 0
Dividing b' by equation (1),
\(( \frac{a}{b})x+y+(\frac{c}{b}) = 0 \)
\(Ax+y=C = 0, where A = \frac{a}{b}, C = \frac{c}{b}\) .....(2)
Thus, eventhough 3 arbitrary constants (a, b, c) are present in (1), they can be considered as 2 constants only, as above (2).
Differentiating (1) with respect to x
a + b \(\\ \frac { dy }{ dx } =0\)
Differentiating again with respect to 'x' we get,
(b) \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\quad [\because b\neq 0]\) .....(3)
This is the differential equation of family of all non - vertical lines in a plane.