Order and degree of differential equations Questions in English

(A) Given the differential equation: $y = x\frac{dy}{dx} + \sqrt{a^2\left(\frac{dy}{dx}\right)^2 + b^2}$.
Subtract $x\frac{dy}{dx}$ from both sides: $y - x\frac{dy}{dx} = \sqrt{a^2\left(\frac{dy}{dx}\right)^2 + b^2}$.
Squaring both sides to eliminate the square root:
$(y - x\frac{dy}{dx})^2 = a^2\left(\frac{dy}{dx}\right)^2 + b^2$.
Expanding the left side:
$y^2 + x^2\left(\frac{dy}{dx}\right)^2 - 2xy\frac{dy}{dx} = a^2\left(\frac{dy}{dx}\right)^2 + b^2$.
The highest derivative present is $\frac{dy}{dx}$,so the order is $1$.
The highest power of the highest derivative after rationalizing the equation is $2$,so the degree is $2$.
Thus,the order and degree are $1, 2$.

(C) To find the order and degree of the differential equation,we must first eliminate the fractional exponent.
Given equation: $\left[ 4 + \left( \frac{dy}{dx} \right)^2 \right]^{2/3} = \frac{d^2y}{dx^2}$
Cube both sides to remove the fraction in the exponent:
$\left[ 4 + \left( \frac{dy}{dx} \right)^2 \right]^2 = \left( \frac{d^2y}{dx^2} \right)^3$
Now,the equation is in the form of a polynomial in terms of derivatives.
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order is $2$.
The power of the highest order derivative after making it a polynomial is $3$,so the degree is $3$.
Thus,the order is $2$ and the degree is $3$.

(C) The order of a differential equation is the highest derivative present,and the degree is the power of the highest derivative when the equation is expressed as a polynomial in derivatives.
Option $(a)$: $x\left( \frac{dy}{dx} \right)^2 - x + a = 0$ is of first order and second degree.
Option $(b)$: $\frac{d^2y}{dx^2} + xy = 0$ is of second order and first degree.
Option $(c)$: $dy + dx = 0$ can be written as $\frac{dy}{dx} + 1 = 0$. This equation involves only the first derivative $\frac{dy}{dx}$ with power $1$. Thus,it is of first order and first degree.
Therefore,the correct option is $(c)$.

(C) The given differential equation is $\frac{d^2y}{dx^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}$.
To find the degree,we must eliminate the radical sign by squaring both sides:
$\left(\frac{d^2y}{dx^2}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2$.
The order of a differential equation is the highest derivative present,which is $2$ here.
The degree is the power of the highest derivative after the equation is made free from radicals and fractions,which is $2$ here.
Thus,the order is $2$ and the degree is $2$.

(A) The given differential equation is $\left( \frac{d^2s}{dt^2} \right)^2 + 3\left( \frac{ds}{dt} \right)^3 + 4 = 0$.
The order of a differential equation is the highest order derivative present in the equation. Here,the highest order derivative is $\frac{d^2s}{dt^2}$,so the order is $2$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{d^2s}{dt^2}$ is $2$.
Therefore,the order is $2$ and the degree is $2$.

(A) The order of a differential equation is the order of the highest derivative present in the equation.
In the given equation $\frac{d^4y}{dx^4} - 4\frac{d^3y}{dx^3} + 8\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 4y = 0$,the highest derivative is $\frac{d^4y}{dx^4}$,which is of order $4$.
Therefore,the order of the differential equation is $4$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives.
Here,the power of the highest derivative $\frac{d^4y}{dx^4}$ is $1$.
Thus,the degree is $1$.
Hence,the order and degree are $4$ and $1$ respectively.

(A) Given the differential equation: $y \left( \frac{dy}{dx} \right) = \frac{x}{\frac{dy}{dx} + \left( \frac{dy}{dx} \right)^3}$.
Multiply both sides by $\left( \frac{dy}{dx} + \left( \frac{dy}{dx} \right)^3 \right)$ to clear the fraction:
$y \left( \frac{dy}{dx} \right) \left( \frac{dy}{dx} + \left( \frac{dy}{dx} \right)^3 \right) = x$.
Expanding this,we get:
$y \left( \frac{dy}{dx} \right)^2 + y \left( \frac{dy}{dx} \right)^4 = x$.
The order of a differential equation is the order of the highest derivative present in the equation.
Here,the highest derivative is $\frac{dy}{dx}$,which is a first-order derivative.
Therefore,the order of the differential equation is $1$.

(C) The given equation is ${x^2} + {y^2} + 2gx + 2fy + c = 0$.
This is the general equation of a circle,which contains $3$ arbitrary constants,namely $g$,$f$,and $c$.
The order of a differential equation is equal to the number of independent arbitrary constants present in its general solution.
Since there are $3$ arbitrary constants,the order of the differential equation is $3$.

(A) The equation of a family of circles of radius $r$ passing through the origin and having its centre on the $y$-axis is given by $(x - 0)^2 + (y - r)^2 = r^2$.
Expanding this,we get $x^2 + y^2 - 2ry + r^2 = r^2$,which simplifies to $x^2 + y^2 - 2ry = 0$.
This equation contains only one arbitrary constant,$r$.
The order of a differential equation is equal to the number of independent arbitrary constants in the general solution of the corresponding family of curves.
Since there is only one arbitrary constant $r$,the order of the differential equation is $1$.

(A) The given solution is $y = a\cos x + b\sin x + c{e^{ - x}}$.
This equation contains $3$ arbitrary constants,namely $a$,$b$,and $c$.
The order of a differential equation is equal to the number of independent arbitrary constants present in its general solution.
Since there are $3$ arbitrary constants,the order of the corresponding differential equation is $3$.

(A) The equation of a circle in the first quadrant that touches both coordinate axes is given by $(x - a)^2 + (y - a)^2 = a^2$,where $a$ is the radius of the circle and acts as an arbitrary constant.
Since there is only one arbitrary constant $a$,the family of curves is a one-parameter family.
The order of the differential equation for a family of curves with $n$ arbitrary constants is $n$.
Therefore,for this family of circles,the order of the differential equation is $1$.

(D) To find the order and degree,we first eliminate the fractional exponent by raising both sides of the equation to the power of $4$.
$\left( \frac{d^2y}{dx^2} \right)^4 = y + \left( \frac{dy}{dx} \right)^2$
The order of a differential equation is the order of the highest derivative present. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{d^2y}{dx^2}$ is $4$.
Thus,the order is $2$ and the degree is $4$.

(B) Given the differential equation: $\frac{d^2y}{dx^2} + \sqrt{1 + \left( \frac{dy}{dx} \right)^3} = 0$
Rearrange the equation to isolate the radical term:
$\frac{d^2y}{dx^2} = - \sqrt{1 + \left( \frac{dy}{dx} \right)^3}$
To eliminate the square root,square both sides of the equation:
$\left( \frac{d^2y}{dx^2} \right)^2 = 1 + \left( \frac{dy}{dx} \right)^3$
The degree of a differential equation is the highest power of the highest-order derivative when the equation is expressed as a polynomial in its derivatives.
Here,the highest-order derivative is $\frac{d^2y}{dx^2}$,and its power is $2$.
Therefore,the degree of the differential equation is $2$.

(D) To find the degree of a differential equation,it must first be expressed as a polynomial in terms of its derivatives.
Given equation: $\left( \frac{d^2y}{dx^2} \right)^3 = \left( 1 + \frac{dy}{dx} \right)^{1/2}$.
To eliminate the fractional exponent,square both sides of the equation:
$\left( \left( \frac{d^2y}{dx^2} \right)^3 \right)^2 = \left( \left( 1 + \frac{dy}{dx} \right)^{1/2} \right)^2$.
This simplifies to: $\left( \frac{d^2y}{dx^2} \right)^6 = 1 + \frac{dy}{dx}$.
The degree of a differential equation is the highest power of the highest order derivative present in the equation after it has been made free from radicals and fractions.
The highest order derivative here is $\frac{d^2y}{dx^2}$,and its power is $6$.
Therefore,the degree of the differential equation is $6$.

(A) Given the differential equation: $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^{1/3} + x^{1/4} = 0$.
To find the degree,we must eliminate the fractional exponents.
Rewrite the equation as: $\frac{d^2y}{dx^2} + x^{1/4} = -\left( \frac{dy}{dx} \right)^{1/3}$.
Cube both sides to remove the exponent $1/3$: $\left( \frac{d^2y}{dx^2} + x^{1/4} \right)^3 = -\frac{dy}{dx}$.
Expanding the left side,the highest derivative term is $\left( \frac{d^2y}{dx^2} \right)^3$.
The order of the highest derivative is $2$.
The degree is the power of the highest derivative after making the equation a polynomial in derivatives,which is $3$.
Thus,the order is $2$ and the degree is $3$.

(A) Given the equation of the family of curves: $y = Ax + A^3$.
Differentiating both sides with respect to $x$,we get: $\frac{dy}{dx} = A$.
Substituting $A = \frac{dy}{dx}$ back into the original equation,we obtain the differential equation: $y = x \left( \frac{dy}{dx} \right) + \left( \frac{dy}{dx} \right)^3$.
The degree of a differential equation is the highest power of the highest-order derivative when the equation is expressed as a polynomial in derivatives.
Here,the highest-order derivative is $\frac{dy}{dx}$,and its highest power is $3$.
Therefore,the degree of the differential equation is $3$.

(C) To find the order and degree of a differential equation,we first express it as a polynomial in terms of its derivatives.
For option $(a)$: $\frac{d^4y}{dx^4} + 8\left(\frac{dy}{dx}\right)^6 + 5y = e^x$. The highest order derivative is $\frac{d^4y}{dx^4}$,so order $= 4$. The power of the highest order derivative is $1$,so degree $= 1$.
For option $(b)$: $5\left(\frac{d^3y}{dx^3}\right)^4 + 8\left(1 + \frac{dy}{dx}\right)^2 + 5y = x^8$. The highest order derivative is $\frac{d^3y}{dx^3}$,so order $= 3$. The power of the highest order derivative is $4$,so degree $= 4$.
For option $(c)$: $\left[1 + \left(\frac{dy}{dx}\right)^3\right]^{2/3} = 4\frac{d^3y}{dx^3}$. Cubing both sides,we get $\left[1 + \left(\frac{dy}{dx}\right)^3\right]^2 = 64\left(\frac{d^3y}{dx^3}\right)^3$. The highest order derivative is $\frac{d^3y}{dx^3}$,so order $= 3$. The power of the highest order derivative is $3$,so degree $= 3$. Here,order $=$ degree.
For option $(d)$: $y - x^2\frac{dy}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}$. Squaring both sides,we get $(y - x^2\frac{dy}{dx})^2 = 1 + \left(\frac{dy}{dx}\right)^2$. The highest order derivative is $\frac{dy}{dx}$,so order $= 1$. The power of the highest order derivative is $2$,so degree $= 2$.

(D) The given differential equation is $\frac{d^2y}{dx^2} + x\frac{dy}{dx} + \sin y + x^2 = 0$.
$1$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.
$2$. The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Since $\sin y$ is present,the equation is non-linear,but the highest derivative $\frac{d^2y}{dx^2}$ has an exponent of $1$. Thus,the degree is $1$.
Therefore,the equation is of order $2$ and degree $1$. Hence,option $(d)$ is correct.

(A) The given differential equation is $x(\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y = x^2$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Here,the power of the highest derivative $\frac{d^2y}{dx^2}$ is $3$.
Therefore,the degree is $3$ and the order is $2$.

(D) The given differential equation is $\left( \frac{d^2y}{dx^2} \right)^5 + 4\frac{\left( \frac{d^2y}{dx^2} \right)^3}{\left( \frac{d^3y}{dx^3} \right)} + \frac{d^3y}{dx^3} = x^2 - 1$.
To find the degree,we must eliminate the fraction by multiplying the entire equation by $\frac{d^3y}{dx^3}$.
Multiplying by $\frac{d^3y}{dx^3}$,we get: $\left( \frac{d^2y}{dx^2} \right)^5 \left( \frac{d^3y}{dx^3} \right) + 4\left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{d^3y}{dx^3} \right)^2 = (x^2 - 1) \left( \frac{d^3y}{dx^3} \right)$.
The highest order derivative present in the equation is $\frac{d^3y}{dx^3}$,so the order $m = 3$.
The degree is the power of the highest order derivative after the equation is made a polynomial in derivatives. The highest power of $\frac{d^3y}{dx^3}$ is $2$.
Thus,$m = 3$ and $n = 2$.

(B) The order of a differential equation is defined as the order of the highest derivative present in the equation.
In option $A$,the highest derivative is $y'$,which is of order $1$.
In option $B$,the highest derivative is $y''$,which is of order $2$.
In option $C$,the highest derivative is $y'''$,which is of order $3$.
In option $D$,the highest derivative is $y'$,which is of order $1$.
Therefore,the second-order differential equation is $y'y'' + y = \sin x$.

(C) The given differential equation is $x\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 + y^2 = 0$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{d^2y}{dx^2}$ is $1$.
Therefore,the order is $2$ and the degree is $1$.

(A) Given the differential equation: $\rho = \frac{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}}}{{\frac{{d^2y}}{{dx^2}}}}$
Multiply both sides by $\frac{{d^2y}}{{dx^2}}$:
$\rho \cdot \frac{{d^2y}}{{dx^2}} = \left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}$
To eliminate the fractional exponent,square both sides:
$\left( \rho \cdot \frac{{d^2y}}{{dx^2}} \right)^2 = \left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3$
The highest order derivative present is $\frac{{d^2y}}{{dx^2}}$,so the order is $2$.
The power of the highest order derivative after rationalizing the equation is $2$,so the degree is $2$.
Therefore,the order and degree are $2$ and $2$ respectively.

(A) The given differential equation is $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 + 6y = 0$.
The degree of a differential equation is defined as the highest power of the highest order derivative when the equation is expressed as a polynomial in derivatives.
Here,the highest order derivative is $\frac{d^2y}{dx^2}$,which is of order $2$.
The power of this highest order derivative is $1$.
Therefore,the degree of the differential equation is $1$.

(D) The given differential equation is $\left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^4 - xy = 0$.
$1$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.
$2$. The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. The highest order derivative is $\frac{d^2y}{dx^2}$,and its power is $3$.
Therefore,the order is $2$ and the degree is $3$.

(A) The given differential equation is $\frac{d^3y}{dx^3} + 2\left[ 1 + \frac{d^2y}{dx^2} \right] = 1$.
Expanding the equation,we get $\frac{d^3y}{dx^3} + 2 + 2\frac{d^2y}{dx^2} = 1$,which simplifies to $\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + 1 = 0$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^3y}{dx^3}$,so the order is $3$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. The power of $\frac{d^3y}{dx^3}$ is $1$.
Thus,the order is $3$ and the degree is $1$.

(B) Given differential equation is ${\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)^{3/4}} = {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^{1/3}}$.
To find the degree,we must eliminate the fractional exponents by raising both sides to the power of the least common multiple of the denominators,which is $12$.
Raising both sides to the power of $12$:
${\left[ {{\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)^{3/4}}} \right]^{12}} = {\left[ {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^{1/3}}} \right]^{12}}$
${\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)^9} = {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^4}$.
The highest order derivative present is $\frac{{{d^2}y}}{{d{x^2}}}$,which is of order $2$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives.
Here,the power of the highest order derivative $\frac{{{d^2}y}}{{d{x^2}}}$ is $4$.
Therefore,the degree is $4$.

(C) The given differential equation is $x \left( \frac{dy}{dx} \right)^3 + 2 \left( \frac{d^2y}{dx^2} \right)^2 + 3y + x = 0$.
The order of a differential equation is the order of the highest derivative present in the equation.
Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives.
The power of the highest derivative $\frac{d^2y}{dx^2}$ is $2$.
Therefore,the order is $2$ and the degree is $2$.

(A) Given the differential equation: $\frac{d^2y}{dx^2} - \sqrt{\frac{dy}{dx} - 3} = x$
Rearrange the equation to isolate the radical term: $\frac{d^2y}{dx^2} - x = \sqrt{\frac{dy}{dx} - 3}$
Squaring both sides to eliminate the square root: $\left(\frac{d^2y}{dx^2} - x\right)^2 = \frac{dy}{dx} - 3$
Expanding the left side: $\left(\frac{d^2y}{dx^2}\right)^2 - 2x\frac{d^2y}{dx^2} + x^2 = \frac{dy}{dx} - 3$
The degree of a differential equation is the highest power of the highest order derivative when the equation is expressed as a polynomial in derivatives.
The highest order derivative is $\frac{d^2y}{dx^2}$,and its highest power is $2$.
Therefore,the degree is $2$.

(A) The general equation of a parabola with its directrix parallel to the $x$-axis is given by $(x - h)^2 = 4a(y - k)$.
Here,$h$,$k$,and $a$ are three independent arbitrary constants.
The order of a differential equation is equal to the number of independent arbitrary constants present in the general equation of the family of curves.
Since there are $3$ arbitrary constants,the order of the differential equation is $3$.

(C) Given the differential equation: $y = 1 + \frac{dy}{dx} + \frac{1}{2!} \left( \frac{dy}{dx} \right)^2 + \frac{1}{3!} \left( \frac{dy}{dx} \right)^3 + \dots$
Let $t = \frac{dy}{dx}$. The series expansion is $y = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \dots$
This is the Taylor series expansion for $e^t$,so $y = e^t$.
Substituting $t = \frac{dy}{dx}$ back,we get $y = e^{\frac{dy}{dx}}$.
Taking the natural logarithm on both sides: $\ln(y) = \frac{dy}{dx}$.
The differential equation is now in the form $\frac{dy}{dx} = \ln(y)$.
The highest order derivative is $\frac{dy}{dx}$,and its power is $1$.
Therefore,the degree of the differential equation is $1$.

(B) differential equation is called linear if the dependent variable and its derivatives occur only in the first degree and are not multiplied together.
In option $A$,$\frac{dy}{dx} + \frac{y}{x} = \log x$ is linear because the dependent variable $y$ and its derivative $\frac{dy}{dx}$ are in the first degree and not multiplied.
In option $B$,$y\frac{dy}{dx} + 4x = 0$,the dependent variable $y$ is multiplied by its derivative $\frac{dy}{dx}$,making it non-linear.
In option $C$,$dx + dy = 0$ can be written as $1 + \frac{dy}{dx} = 0$,which is linear.
In option $D$,$\frac{dy}{dx} = \cos x$ is linear as the derivative is in the first degree.
Therefore,the correct option is $B$.

(D) differential equation is linear if the dependent variable $y$ and its derivatives appear only in the first degree and are not multiplied together.
For option $(d)$,we have $x\frac{dy}{dx} + \frac{3}{dy/dx} = y^2$.
Multiplying both sides by $\frac{dy}{dx}$,we get $x\left(\frac{dy}{dx}\right)^2 - y^2\frac{dy}{dx} + 3 = 0$.
Since the degree of the derivative $\frac{dy}{dx}$ is $2$,this differential equation is non-linear.

(B) Given the differential equation: $3\frac{d^2y}{dx^2} = \left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^{3/2}$
To find the degree,we must eliminate the fractional exponent by squaring both sides:
$\left( 3\frac{d^2y}{dx^2} \right)^2 = \left[ \left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^{3/2} \right]^2$
$9\left( \frac{d^2y}{dx^2} \right)^2 = \left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^3$
The highest order derivative present in the equation is $\frac{d^2y}{dx^2}$,which is of order $2$.
The power (exponent) of the highest order derivative after making the equation a polynomial in derivatives is $2$.
Therefore,the degree of the differential equation is $2$.

(D) Given curve is ${y^2} = 2c(x + \sqrt{c}).$
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2c$,which implies $c = y \frac{dy}{dx}.$
Substituting $c$ into the original equation:
${y^2} = 2 \left( y \frac{dy}{dx} \right) \left( x + \sqrt{y \frac{dy}{dx}} \right).$
Rearranging the terms:
$\frac{y}{2(dy/dx)} - x = \sqrt{y \frac{dy}{dx}}.$
Squaring both sides:
$\left( \frac{y}{2(dy/dx)} - x \right)^2 = y \frac{dy}{dx}.$
Multiplying by $4(dy/dx)^2$:
$(y - 2x(dy/dx))^2 = 4y(dy/dx)^3.$
Expanding the square:
${y^2} - 4xy \frac{dy}{dx} + 4{x^2} \left( \frac{dy}{dx} \right)^2 = 4y \left( \frac{dy}{dx} \right)^3.$
Rearranging into the standard form:
$4y \left( \frac{dy}{dx} \right)^3 - 4{x^2} \left( \frac{dy}{dx} \right)^2 + 4xy \frac{dy}{dx} - {y^2} = 0.$
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$. The highest power of the highest order derivative is $3$,so the degree is $3$.
Thus,the differential equation is of order $1$ and degree $3$.

(B) The given general solution is $y = C_1 e^{2x + C_2} + C_3 e^x + C_4 \sin(x + C_5)$.
We can simplify the expression as follows:
$y = C_1 e^{C_2} e^{2x} + C_3 e^x + C_4 (\sin x \cos C_5 + \cos x \sin C_5)$
Let $A = C_1 e^{C_2}$,$B = C_4 \cos C_5$,and $D = C_4 \sin C_5$.
Substituting these constants,we get:
$y = A e^{2x} + C_3 e^x + B \sin x + D \cos x$
In this expression,there are $4$ independent arbitrary constants $(A, C_3, B, D)$.
The order of a differential equation is equal to the number of independent arbitrary constants present in its general solution.
Therefore,the order of the differential equation is $4$.

(C) To find the order and degree of the differential equation,we must first eliminate the fractional exponent by raising both sides to the power of $3$.
Given equation: ${\left( {1 + 3\frac{{dy}}{{dx}}} \right)^{\frac{2}{3}}} = 4\frac{{{d^3}y}}{{d{x^3}}}$
Cubing both sides:
${\left( {1 + 3\frac{{dy}}{{dx}}} \right)^2} = {\left( {4\frac{{{d^3}y}}{{d{x^3}}}} \right)^3}$
Now,the equation is a polynomial in terms of derivatives.
The order of a differential equation is the highest derivative present,which is $\frac{{{d^3}y}}{{d{x^3}}}$,so the order is $3$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{{{d^3}y}}{{d{x^3}}}$ is $3$.
Thus,the order is $3$ and the degree is $3$.

(D) The given differential equation is $\left( \frac{d^2y}{dx^2} \right)^2 - \left( \frac{dy}{dx} \right)^{1/2} = y^3$.
To find the degree,we must first express the equation as a polynomial in terms of its derivatives by removing fractional exponents.
Rearranging the equation,we get $\left( \frac{d^2y}{dx^2} \right)^2 - y^3 = \left( \frac{dy}{dx} \right)^{1/2}$.
Squaring both sides to eliminate the fractional exponent,we obtain $\left( \left( \frac{d^2y}{dx^2} \right)^2 - y^3 \right)^2 = \frac{dy}{dx}$.
Expanding the left side,the highest order derivative present is $\frac{d^2y}{dx^2}$,which is of order $2$.
The power to which the highest order derivative is raised in this polynomial form is $2 \times 2 = 4$.
Therefore,the degree of the differential equation is $4$.

(A) The general equation of a parabola with its axis of symmetry as the $x$-axis is given by $y^2 = 4a(x - h)$.
Here,$a$ and $h$ are two arbitrary constants.
To form the differential equation,we differentiate with respect to $x$:
$2y \frac{dy}{dx} = 4a$
$y \frac{dy}{dx} = 2a$
Differentiating again with respect to $x$:
$y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0$
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order is $2$.
The power of the highest order derivative is $1$,so the degree is $1$.
Thus,the order and degree are $(2, 1)$.

(D) Given equation: $y = a(1 - e^{-x/a})$
$y = a - ae^{-x/a}$
$y - a = -ae^{-x/a}$
Divide by $-a$: $\frac{a - y}{a} = e^{-x/a}$
Taking natural logarithm on both sides: $\ln(\frac{a - y}{a}) = -\frac{x}{a}$
Since $a$ is a parameter,we differentiate with respect to $x$ to eliminate it.
However,the equation contains an exponential term involving the parameter $a$ in the exponent,which cannot be eliminated to form a polynomial differential equation in terms of $y'$ and $y''$.
Specifically,the presence of the parameter $a$ in the exponent $e^{-x/a}$ makes it impossible to express the differential equation as a polynomial in derivatives.
Therefore,the degree of this differential equation is not defined.

(C) Given the differential equation: $(\frac{d^3y}{dx^3}+y\frac{dy}{dx})^{\frac{7}{5}} =x^3\frac{d^2y}{dx^2}$.
To find the degree,we must eliminate the fractional exponent by raising both sides to the power of $5$:
$(\frac{d^3y}{dx^3}+y\frac{dy}{dx})^7 = (x^3\frac{d^2y}{dx^2})^5$.
The order $m$ is the highest derivative present,which is $\frac{d^3y}{dx^3}$,so $m = 3$.
The degree $n$ is the power of the highest derivative after the equation is expressed as a polynomial in derivatives,which is $7$,so $n = 7$.
Therefore,$m + n = 3 + 7 = 10$.

(B) Given the differential equation: $\left( \frac{d^3y}{dx^3} \right)^{\frac{2}{3}} + 4 - 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} = 0$.
To find the degree,we must first eliminate the fractional exponent of the highest order derivative.
Rearrange the equation: $\left( \frac{d^3y}{dx^3} \right)^{\frac{2}{3}} = 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4$.
Cube both sides to remove the fractional power: $\left( \left( \frac{d^3y}{dx^3} \right)^{\frac{2}{3}} \right)^3 = \left( 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4 \right)^3$.
This simplifies to: $\left( \frac{d^3y}{dx^3} \right)^2 = \left( 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4 \right)^3$.
The highest order derivative present is $\frac{d^3y}{dx^3}$,which is of order $3$.
The power (exponent) of the highest order derivative after making the equation a polynomial in derivatives is $2$.
Therefore,the degree of the differential equation is $2$.

(D) The degree of a differential equation is defined only when it is a polynomial equation in terms of its derivatives.
In the given equation,$\left( \frac{d^3y}{dx^3} \right)^2 + 4\left( \frac{dy}{dx} \right)^3 = 3\sin \left( \frac{d^2y}{dx^2} \right)$,the term $\sin \left( \frac{d^2y}{dx^2} \right)$ involves a transcendental function of a derivative.
Since the equation cannot be expressed as a polynomial in terms of its derivatives,the degree is not defined.

(B) differential equation is said to be linear if the dependent variable $y$ and its derivatives occur only in the first degree and are not multiplied together.
In the given equation $\frac{d^3y}{dx^3}-5y \frac{dy}{dx}+xy=0$,the term $y \frac{dy}{dx}$ involves the product of the dependent variable $y$ and its derivative $\frac{dy}{dx}$.
Because of this product,the equation is non-linear.
The order of the highest derivative $\frac{d^3y}{dx^3}$ is $3$.
The degree of the highest derivative is $1$.
Therefore,it is a non-linear differential equation of order $3$ and degree $1$.

(D) The given differential equation is $\frac{d^2y}{dx^2} = \cos \left( \frac{dy}{dx} \right) + xy$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in its derivatives.
In this equation,the term $\cos \left( \frac{dy}{dx} \right)$ involves a transcendental function of the derivative $\frac{dy}{dx}$,which means the equation cannot be expressed as a polynomial in its derivatives.
Therefore,the degree of the differential equation is not defined.

(D) For Statement $1$:
Consider the first differential equation: $\frac{dy}{dx} + y^2 = x$. The highest order derivative is $\frac{dy}{dx}$ (order $1$),and its power is $1$. Thus,the degree is $1$.
Consider the second differential equation: $\frac{d^2y}{dx^2} + y = \sin x$. The highest order derivative is $\frac{d^2y}{dx^2}$ (order $2$),and its power is $1$. Thus,the degree is $1$.
Since both equations have a degree of $1$,Statement $1$ is true.
For Statement $2$:
This is the standard definition of the degree of a differential equation. It is a polynomial in derivatives,and the degree is the highest power of the highest order derivative. Thus,Statement $2$ is true.
Conclusion:
Statement $2$ provides the definition used to determine the degrees in Statement $1$,making it the correct explanation. Therefore,the correct option is $D$.

(A) The given differential equation is $\frac{dy}{dx} - \cos x = 0$.
The highest order derivative present in the equation is $\frac{dy}{dx}$,which is a first-order derivative. Therefore,the order of the differential equation is $1$.
The equation is a polynomial in terms of the derivative $\frac{dy}{dx}$. The highest power of the highest order derivative $\frac{dy}{dx}$ is $1$. Therefore,the degree of the differential equation is $1$.

(A) The given differential equation is $x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$.
$1$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^{2} y}{d x^{2}}$,so the order is $2$.
$2$. The degree of a differential equation is the power of the highest order derivative,provided the equation is a polynomial in derivatives. Here,the highest order derivative $\frac{d^{2} y}{d x^{2}}$ has an exponent of $1$. Since the equation is a polynomial in terms of its derivatives,the degree is $1$.
Therefore,the order is $2$ and the degree is $1$.

(C) The given differential equation is $y^{\prime \prime \prime} + y^2 + e^{y^{\prime}} = 0$.
The highest order derivative present in the equation is $y^{\prime \prime \prime}$,which represents the third derivative.
Therefore,the order of the differential equation is $3$.
$A$ differential equation is said to have a defined degree if it can be expressed as a polynomial in its derivatives.
In this equation,the term $e^{y^{\prime}}$ involves a transcendental function of the derivative $y^{\prime}$,which means the equation cannot be expressed as a polynomial in its derivatives.
Thus,the degree of this differential equation is not defined.