Order and degree of differential equations Questions in English

(C) The given differential equation is $\frac{d^4y}{dx^4} + \sin(y''') = 0$.
The highest order derivative present in the equation is $\frac{d^4y}{dx^4}$,which is of order $4$.
$A$ differential equation is said to have a defined degree only if it can be expressed as a polynomial in its derivatives.
In this equation,the term $\sin(y''')$ involves a transcendental function of a derivative,which means the equation cannot be written as a polynomial in its derivatives.
Therefore,the order is $4$ and the degree is not defined.

(A) The given differential equation is $y^{\prime} + 5y = 0$.
The highest order derivative present in the differential equation is $y^{\prime} = \frac{dy}{dx}$.
Since the highest derivative is of the first order,the order of the differential equation is $1$.
The power of the highest order derivative $y^{\prime}$ is $1$.
Therefore,the degree of the differential equation is $1$.

(A) The given differential equation is $\left(\frac{ds}{dt}\right)^{4} + 3s \frac{d^{2}s}{dt^{2}} = 0$.
The highest order derivative present in the equation is $\frac{d^{2}s}{dt^{2}}$.
Since the highest order derivative is of the second order,the order of the differential equation 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 terms of its derivatives.
The highest order derivative $\frac{d^{2}s}{dt^{2}}$ has an exponent of $1$.
Therefore,the degree of the differential equation is $1$.

(C) The given differential equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^{2} y}{d x^{2}}$,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 $\cos \left(\frac{d y}{d x}\right)$ involves a transcendental function of the derivative,which means the equation cannot be expressed as a polynomial in terms of its derivatives.
Therefore,the degree of this differential equation is not defined.

(A) The given differential equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{2} = \cos 3x + \sin 3x$.
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 $2$.
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 the given equation,the highest order derivative $\frac{d^{2} y}{d x^{2}}$ is raised to the power of $2$.
Therefore,the degree is $2$.

(A) The given differential equation is $(\frac{d^3y}{dx^3})^2 + (\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y^5 = 0$.
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 order derivative,provided the equation is a polynomial in its derivatives. Here,the highest order derivative $\frac{d^3y}{dx^3}$ is raised to the power of $2$. Therefore,the degree is $2$.

(A) The given differential equation is $y''' + 2y'' + y' = 0$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $y'''$,which is the third derivative. Thus,the order is $3$.
The degree of a differential equation is the power of the highest order derivative,provided the equation is a polynomial in derivatives. Here,the power of $y'''$ is $1$. Thus,the degree is $1$.

(A) The given differential equation is $y^{\prime} + y = e^{x}$.
The highest order derivative present in the equation is $y^{\prime} = \frac{dy}{dx}$,which is of order $1$.
Since the differential equation is a polynomial in terms of its derivatives,the degree is the highest power of the highest order derivative.
The power of $y^{\prime}$ is $1$. Therefore,the degree is $1$.
Thus,the order is $1$ and the degree is $1$.

(A) The given differential equation is $y'' + (y')^2 + 2y = 0$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $y''$,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. Here,the highest order derivative $y''$ has an exponent of $1$.
Thus,the order is $2$ and the degree is $1$.

(A) The given differential equation is $y^{\prime \prime} + 2y^{\prime} + \sin(y) = 0$.
The highest order derivative present in the equation is $y^{\prime \prime}$,which represents the second derivative. Therefore,the order of the differential equation is $2$.
$A$ differential equation is a polynomial in derivatives if the derivatives are not arguments of transcendental functions like $\sin$,$\cos$,$e^x$,etc. In this equation,the term $\sin(y)$ involves $y$,but the derivatives $y^{\prime \prime}$ and $y^{\prime}$ are not inside a transcendental function. Thus,the equation is a polynomial in its derivatives.
The highest power of the highest order derivative $y^{\prime \prime}$ is $1$. Therefore,the degree of the differential equation is $1$.

(B) The given differential equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}+\sin \left(\frac{d y}{d x}\right)+1=0$.
The degree of a differential equation is defined only when it can be expressed as a polynomial in its derivatives.
In this equation,the term $\sin \left(\frac{d y}{d x}\right)$ involves a derivative as an argument of a transcendental function,which prevents the equation from being a polynomial in its derivatives.
Therefore,the degree of this differential equation is not defined.
Hence,the correct answer is $B$.

(C) The given differential equation is $2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0$.
The order of a differential equation is defined as the order of the highest order derivative present in the equation.
In this equation,the highest order derivative is $\frac{d^{2} y}{d x^{2}}$,which is a second-order derivative.
Therefore,the order of the differential equation is $2$.
Hence,the correct answer is $C$.

(D) We know that the number of arbitrary constants in the general solution of a differential equation of order $n$ is equal to its order $n$.
Since the given differential equation is of the fourth order,its order $n = 4$.
Therefore,the number of arbitrary constants in its general solution is $4$.
Hence,the correct answer is $D$.

(A) general solution of a differential equation of order $n$ contains $n$ arbitrary constants.
$A$ particular solution is obtained by assigning specific values to these arbitrary constants.
Therefore,a particular solution contains no arbitrary constants.
Hence,the correct answer is $A$.

(A) The given differential equation is:
$\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$
$1$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^{2} y}{d x^{2}}$,which is of order $2$. Therefore,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 $\frac{d^{2} y}{d x^{2}}$,and its exponent is $1$. Therefore,the degree is $1$.
Thus,the order is $2$ and the degree is $1$.

(A) The given differential equation is:
$\left(\frac{dy}{dx}\right)^{3}-4\left(\frac{dy}{dx}\right)^{2}+7y=\sin x$
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 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 power of the first-order derivative $\frac{dy}{dx}$ is $3$.
Therefore,the degree of the differential equation is $3$.

(C) The given differential equation is:
$\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0$
The order of a differential equation is the order of the highest derivative present in the equation.
Here,the highest derivative is $\frac{d^{4} y}{d x^{4}}$,so the order is $4$.
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.
Since the term $\sin \left(\frac{d^{3} y}{d x^{3}}\right)$ involves a transcendental function of a derivative,the equation cannot be expressed as a polynomial in its derivatives.
Therefore,the degree of this differential equation is not defined.

(A) The equation of a tangent line to the parabola $x^2 = 4y$ with slope $m$ is given by $y = mx - am^2$. Here,$a = 1$,so the equation is $y = mx - m^2$.
To find the differential equation,we differentiate with respect to $x$:
$\frac{dy}{dx} = m$.
Substituting $m = \frac{dy}{dx}$ into the original equation,we get:
$y = x(\frac{dy}{dx}) - (\frac{dy}{dx})^2$.
Rearranging the terms,we have $(\frac{dy}{dx})^2 - x(\frac{dy}{dx}) + y = 0$.
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$.
The power of the highest order derivative is $2$,so the degree is $2$.

(D) Given differential equation is $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[5]{\frac{dy}{d x}-5}$.
To find the order and degree,we first eliminate the radicals by raising both sides to the power of $10$ (the least common multiple of $2$ and $5$):
$(\frac{d^2 y}{d x^2})^{1/2} = (\frac{dy}{d x}-5)^{1/5}$
$(\frac{d^2 y}{d x^2})^{10/2} = (\frac{dy}{d x}-5)^{10/5}$
$(\frac{d^2 y}{d x^2})^5 = (\frac{dy}{d x}-5)^2$.
The order of the differential equation is the highest derivative present,which is $2$.
The degree of the differential equation is the power of the highest derivative after making the equation a polynomial in derivatives,which is $5$.
Therefore,the sum of the degree and order is $2 + 5 = 7$.

(D) The given differential equation is $\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)$.
$A$ differential equation is said to be a polynomial equation in its derivatives if it can be expressed as a polynomial in terms of its derivatives.
In the given equation,the term $\log \left(\frac{d^2 y}{d x^2}\right)$ involves the second-order derivative inside a logarithmic function,which means the equation cannot be expressed as a polynomial in terms of its derivatives.
Therefore,the degree of this differential equation is not defined.

(A) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4\frac{dy}{dx} - 7x = 0$.
Rearranging the terms,we get $\sqrt{\frac{dy}{dx}} = 4\frac{dy}{dx} + 7x$.
Squaring both sides to eliminate the square root,we get $\frac{dy}{dx} = (4\frac{dy}{dx} + 7x)^2$.
Expanding the right side,$\frac{dy}{dx} = 16(\frac{dy}{dx})^2 + 56x\frac{dy}{dx} + 49x^2$.
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$.
The highest power of the highest order derivative after making the equation a polynomial in derivatives is $2$.
Therefore,the order is $1$ and the degree is $2$.

(A) Given the equation $Ax^2 + By^2 = 1$.
Since there are $2$ arbitrary constants ($A$ and $B$),we differentiate the equation twice.
First differentiation with respect to $x$: $2Ax + 2Byy' = 0$,which simplifies to $Ax + Byy' = 0$.
Second differentiation with respect to $x$: $A + B(y')^2 + Byy'' = 0$.
From the first derivative,$A = -Byy'/x$.
Substituting $A$ into the second derivative: $-Byy'/x + B(y')^2 + Byy'' = 0$.
Dividing by $B$ (assuming $B \neq 0$): $-yy'/x + (y')^2 + yy'' = 0$.
Multiplying by $x$: $-yy' + x(y')^2 + xyy'' = 0$.
The highest order derivative present is $y''$,so the order is $2$.
The power of the highest order derivative $y''$ is $1$,so the degree is $1$.

(A) Given differential equation is $3 - (\frac{d^3 y}{d x^3})^{\frac{7}{3}} = (\frac{dy}{d x})^5$.
To find the degree,we must eliminate the fractional exponent.
Rearrange the equation: $3 - (\frac{dy}{d x})^5 = (\frac{d^3 y}{d x^3})^{\frac{7}{3}}$.
Raise both sides to the power of $3$: $(3 - (\frac{dy}{d x})^5)^3 = (\frac{d^3 y}{d x^3})^7$.
The highest order derivative present is $\frac{d^3 y}{d x^3}$,so the order is $3$.
The power of the highest order derivative after making the equation free from radicals and fractions is $7$,so the degree is $7$.

(A) Given the differential equation: $\left[\frac{1+\left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2y}{dx^2}\right)^{\frac{3}{2}}}\right]^2 = kx$
First,simplify the expression by removing the fractional exponent:
$\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^2}{\left(\frac{d^2y}{dx^2}\right)^3} = kx$
Now,multiply both sides by $\left(\frac{d^2y}{dx^2}\right)^3$:
$\left[1+\left(\frac{dy}{dx}\right)^2\right]^2 = kx \left(\frac{d^2y}{dx^2}\right)^3$
The order of a differential equation is the highest derivative present,which is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree is the power of the highest derivative when the equation is expressed as a polynomial in derivatives,which is $3$.
Therefore,the order is $2$ and the degree is $3$.

(B) Given the differential equation: $\left(\frac{d^2 y}{dx^2}\right)^5 + 4 \frac{\left(\frac{d^2 y}{dx^2}\right)^5}{\frac{d^3 y}{dx^3}} + \frac{d^3 y}{dx^3} = \sin x$.
Multiply the entire equation by $\frac{d^3 y}{dx^3}$ to eliminate the fraction:
$\left(\frac{d^2 y}{dx^2}\right)^5 \cdot \frac{d^3 y}{dx^3} + 4\left(\frac{d^2 y}{dx^2}\right)^5 + \left(\frac{d^3 y}{dx^3}\right)^2 = \sin x \cdot \frac{d^3 y}{dx^3}$.
The highest order derivative present is $\frac{d^3 y}{dx^3}$,so the order $m = 3$.
The highest power of the highest order derivative is $2$,so the degree $n = 2$.
Therefore,$m^2 + n^2 = 3^2 + 2^2 = 9 + 4 = 13$.

(B) Given the general solution: $y = (c_1 + c_2) \cos (x + c_3) - c_4 e^{x + c_5}$.
We can simplify the expression by grouping the constants:
Let $A = c_1 + c_2$ and $B = c_4 e^{c_5}$.
Then the equation becomes $y = A \cos (x + c_3) - B e^x$.
Expanding the cosine term: $y = A (\cos x \cos c_3 - \sin x \sin c_3) - B e^x$.
$y = (A \cos c_3) \cos x - (A \sin c_3) \sin x - B e^x$.
Let $K_1 = A \cos c_3$,$K_2 = -A \sin c_3$,and $K_3 = -B$.
Thus,$y = K_1 \cos x + K_2 \sin x + K_3 e^x$.
There are $3$ independent arbitrary constants $(K_1, K_2, K_3)$.
The order of a differential equation is equal to the number of independent arbitrary constants in its general solution.
Therefore,the order of the differential equation is $3$.

(B) Given the differential equation: $\left[1-\left(\frac{dy}{dx}\right)^2\right]^{5/2} = 8 \frac{d^2y}{dx^2}$.
To find the degree,we must eliminate the fractional exponent by squaring both sides:
$\left[\left[1-\left(\frac{dy}{dx}\right)^2\right]^{5/2}\right]^2 = \left[8 \frac{d^2y}{dx^2}\right]^2$
$\left[1-\left(\frac{dy}{dx}\right)^2\right]^5 = 64 \left(\frac{d^2y}{dx^2}\right)^2$.
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order is $2$.
The power to which the highest order derivative is raised after rationalizing the equation is $2$,so the degree is $2$.

(C) Given the differential equation: $\left(1+\frac{dy}{dx}\right)^{\frac{1}{3}}=\left(\frac{d^2y}{dx^2}\right)^{\frac{1}{2}}$
To eliminate the fractional exponents,we raise both sides to the power of $6$ (the least common multiple of $2$ and $3$):
$\left(\left(1+\frac{dy}{dx}\right)^{\frac{1}{3}}\right)^6 = \left(\left(\frac{d^2y}{dx^2}\right)^{\frac{1}{2}}\right)^6$
$\left(1+\frac{dy}{dx}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^3$
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order is $2$.
The power to which the highest order derivative is raised is $3$,so the degree is $3$.
Thus,the order and degree are $2$ and $3$ respectively.

(A) Given the family of curves: $y^2 = 2C(x + \sqrt{C}) \quad \dots (i)$
Differentiating with respect to $x$:
$2y \frac{dy}{dx} = 2C$
$\Rightarrow C = y \frac{dy}{dx} \quad \dots (ii)$
Substitute the value of $C$ from $(ii)$ into $(i)$:
$y^2 = 2 \left( y \frac{dy}{dx} \right) \left( x + \sqrt{y \frac{dy}{dx}} \right)$
$y = 2 \frac{dy}{dx} \left( x + \sqrt{y \frac{dy}{dx}} \right)$
$y - 2x \frac{dy}{dx} = 2 \frac{dy}{dx} \sqrt{y \frac{dy}{dx}}$
Squaring both sides:
$(y - 2x \frac{dy}{dx})^2 = 4 \left( \frac{dy}{dx} \right)^2 \left( y \frac{dy}{dx} \right)$
$(y - 2x \frac{dy}{dx})^2 = 4y \left( \frac{dy}{dx} \right)^3$
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$.
The power of the highest order derivative is $3$,so the degree is $3$.
Thus,the order and degree are $1$ and $3$ respectively.

(C) Given the differential equation $y = \frac{dp}{dx} + \sqrt{a^2 p^2 - b^2}$,where $p = \frac{dy}{dx}$.
Substituting $p = \frac{dy}{dx}$,we get $\frac{dp}{dx} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d^2y}{dx^2}$.
Thus,the equation becomes $y = \frac{d^2y}{dx^2} + \sqrt{a^2(\frac{dy}{dx})^2 - b^2}$.
Rearranging the terms,we have $y - \frac{d^2y}{dx^2} = \sqrt{a^2(\frac{dy}{dx})^2 - b^2}$.
Squaring both sides to eliminate the square root,we get $(y - \frac{d^2y}{dx^2})^2 = a^2(\frac{dy}{dx})^2 - b^2$.
Expanding the left side,$y^2 + (\frac{d^2y}{dx^2})^2 - 2y(\frac{d^2y}{dx^2}) = a^2(\frac{dy}{dx})^2 - b^2$.
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order $m = 2$.
The highest power of the highest order derivative is $2$,so the degree $n = 2$.
Therefore,$m + n = 2 + 2 = 4$.

(B) Given the differential equation: $\left(\frac{d^2 y}{d x^2}\right)^5+4 \frac{\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d^3 y}{d x^3}\right)}+\left(\frac{d^3 y}{d x^3}\right)=x^2-1$.
Multiply the entire equation by $\left(\frac{d^3 y}{d x^3}\right)$ to eliminate the fraction:
$\left(\frac{d^3 y}{d x^3}\right) \left(\frac{d^2 y}{d x^2}\right)^5 + 4 \left(\frac{d^2 y}{d x^2}\right) + \left(\frac{d^3 y}{d x^3}\right)^2 = (x^2-1) \left(\frac{d^3 y}{d x^3}\right)$.
The highest order derivative present is $\frac{d^3 y}{d x^3}$,so the order $m = 3$.
The highest power of the highest order derivative after clearing fractions is $2$,so the degree $n = 2$.
Therefore,$m=3$ and $n=2$.

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

(B) Given differential equation is $\sqrt{\frac{dy}{dx}}-4 \frac{dy}{dx}-7x=0$.
Rearranging the terms,we get $\sqrt{\frac{dy}{dx}}=4 \frac{dy}{dx}+7x$.
Squaring both sides to eliminate the radical,we obtain $\frac{dy}{dx} = (4 \frac{dy}{dx} + 7x)^2$.
Expanding the right side,we get $\frac{dy}{dx} = 16(\frac{dy}{dx})^2 + 56x(\frac{dy}{dx}) + 49x^2$.
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$.
The highest power of the highest order derivative is $2$,so the degree is $2$.
Thus,the order is $1$ and the degree is $2$.

(C) The degree of a differential equation is defined as the power of the highest ordered derivative,provided the equation can be expressed as a polynomial in terms of its derivatives.
In the given equation $e^{\frac{dy}{dx}} + (\frac{dy}{dx})^3 = x$,the term $e^{\frac{dy}{dx}}$ involves the derivative in the exponent,which means it cannot be expressed as a polynomial in terms of $\frac{dy}{dx}$.
Therefore,the degree of this differential equation is not defined.

(B) Given differential equation is $\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{\frac{7}{3}}=7 \frac{d^{2}y}{dx^{2}}$.
To find the degree,we must eliminate the fractional exponent by raising both sides to the power of $3$:
$\left(\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{\frac{7}{3}}\right)^{3} = \left(7 \frac{d^{2}y}{dx^{2}}\right)^{3}$.
This simplifies to $\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{7} = 343 \left(\frac{d^{2}y}{dx^{2}}\right)^{3}$.
The highest order derivative present is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.
The power of the highest order derivative after making the equation a polynomial in derivatives is $3$,so the degree is $3$.
Thus,the order and degree are $2$ and $3$ respectively.

(A) Given differential equation is $\sqrt{1+\frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{3/2}$.
Squaring both sides,we get $1+\frac{1}{(\frac{dy}{dx})^2} = (\frac{d^2y}{dx^2})^3$.
Multiplying both sides by $(\frac{dy}{dx})^2$,we get $(\frac{dy}{dx})^2 + 1 = (\frac{d^2y}{dx^2})^3 (\frac{dy}{dx})^2$.
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 the equation free from radicals and fractions is $3$,so the degree is $3$.
Thus,the order and degree are $2$ and $3$ respectively.

(A) Given differential equation is $\left[1+\frac{1}{(\frac{dy}{dx})^{2}}\right]^{\frac{5}{3}}=5 \frac{d^{2}y}{dx^{2}}$.
To find the order and degree,we first eliminate the fractional exponent by raising both sides to the power of $3$:
$\left[1+\frac{1}{(\frac{dy}{dx})^{2}}\right]^{5} = (5 \frac{d^{2}y}{dx^{2}})^{3}$
$\left[\frac{(\frac{dy}{dx})^{2}+1}{(\frac{dy}{dx})^{2}}\right]^{5} = 125 (\frac{d^{2}y}{dx^{2}})^{3}$
$((\frac{dy}{dx})^{2}+1)^{5} = 125 (\frac{d^{2}y}{dx^{2}})^{3} (\frac{dy}{dx})^{10}$
The highest order derivative present is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.
The power of the highest order derivative after making the equation a polynomial in derivatives is $3$.
Thus,the order is $2$ and the degree is $3$.

(A) Given the differential equation: $y = px + \sqrt{a^{2}p^{2} + b^{2}}$,where $p = \frac{dy}{dx}$.
Rearranging the terms,we get: $y - px = \sqrt{a^{2}p^{2} + b^{2}}$.
Squaring both sides:
$(y - px)^{2} = a^{2}p^{2} + b^{2}$
$y^{2} - 2pxy + p^{2}x^{2} = a^{2}p^{2} + b^{2}$
Substituting $p = \frac{dy}{dx}$:
$y^{2} - 2xy\left(\frac{dy}{dx}\right) + x^{2}\left(\frac{dy}{dx}\right)^{2} = a^{2}\left(\frac{dy}{dx}\right)^{2} + b^{2}$
The highest derivative present is $\frac{dy}{dx}$,so the order is $1$.
The highest power of the highest derivative is $2$,so the degree is $2$.
Therefore,the order and degree are $1$ and $2$ respectively.

(A) Given the differential equation: $\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{\frac{7}{3}}=7\left(\frac{d^{2}y}{dx^{2}}\right)$.
To find the degree,we must eliminate the fractional exponent. Raise both sides to the power of $3$:
$\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{7} = 7^{3}\left(\frac{d^{2}y}{dx^{2}}\right)^{3}$.
The order of a differential equation is the highest derivative present,which is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.
The degree is the power of the highest order derivative after the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{d^{2}y}{dx^{2}}$ is $3$.
Thus,the order is $2$ and the degree is $3$.

(C) The equation of a circle that lies in the first quadrant and touches both the axes is given by $(x - a)^2 + (y - a)^2 = a^2$,where $a$ is the radius of the circle.
This simplifies to $x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = a^2$,which is $x^2 + y^2 - 2ax - 2ay + a^2 = 0$.
Here,there is only one arbitrary constant,$a$.
The order of the differential equation is equal to the number of independent arbitrary constants in the general equation of the family of curves.
Since there is only $1$ arbitrary constant,the order of the differential equation is $1$.

(B) The given differential equation is $\left[1+\left(\frac{dy}{dx}\right)^3\right]^{\frac{7}{3}}=7\left(\frac{d^2y}{dx^2}\right)$.
To find the degree,we must eliminate the fractional exponent by raising both sides to the power of $3$:
$\left[\left[1+\left(\frac{dy}{dx}\right)^3\right]^{\frac{7}{3}}\right]^3 = \left[7\left(\frac{d^2y}{dx^2}\right)\right]^3$
$\Rightarrow \left[1+\left(\frac{dy}{dx}\right)^3\right]^7 = 343\left(\frac{d^2y}{dx^2}\right)^3$.
The order of a differential equation is the highest derivative present,which is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree is the power of the highest derivative after the equation is made a polynomial in derivatives,which is $3$.
Thus,the degree is $3$ and the order is $2$.

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

(A) Given differential equation is $\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1-\left(\frac{d y}{d x}\right)^{4}}$.
To find the order and degree,we first eliminate the radical sign by cubing both sides:
$\left(\frac{d^{2} y}{d x^{2}}\right)^{3} = 1 - \left(\frac{d y}{d x}\right)^{4}$.
The order of a differential equation is the order of the highest derivative present. Here,the highest derivative is $\frac{d^{2} y}{d x^{2}}$,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 $\frac{d^{2} y}{d x^{2}}$ is $3$.
Therefore,the order is $2$ and the degree is $3$.

(D) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0$.
To find the degree,we must eliminate the radical sign.
Rewrite the equation as $\sqrt{\frac{dy}{dx}} = 4 \frac{dy}{dx} + 7x$.
Squaring both sides,we get:
$(\frac{dy}{dx}) = (4 \frac{dy}{dx} + 7x)^2$.
$(\frac{dy}{dx}) = 16(\frac{dy}{dx})^2 + 49x^2 + 56x \frac{dy}{dx}$.
Rearranging the terms,we get $16(\frac{dy}{dx})^2 + (56x - 1)\frac{dy}{dx} + 49x^2 = 0$.
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$.
The highest power of the highest order derivative is $2$,so the degree is $2$.
Thus,the order is $1$ and the degree is $2$.

(B) Given the solution: $y=(C_1+C_2) e^x+C_3 e^{x+C_4}$
We can simplify the expression as follows:
$y = (C_1+C_2) e^x + C_3 e^x \cdot e^{C_4}$
Let $A = C_1+C_2$ and $B = C_3 e^{C_4}$.
Then the equation becomes $y = Ae^x + Be^x = (A+B) e^x$.
Let $D = A+B$,where $D$ is a single arbitrary constant.
Thus,the equation simplifies to $y = De^x$.
Since the final form of the solution contains only one arbitrary constant,the order of the corresponding differential equation is $1$.

(A) Given the differential equation: $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left(\frac{d y}{d x}\right)^4+2}$.
To find the order and degree,we first remove the radicals by raising both sides to the power of $6$ (the least common multiple of $2$ and $3$):
$\left(\left(\frac{d^2 y}{d x^2}\right)^{1/2}\right)^6 = \left(\left(\left(\frac{d y}{d x}\right)^4+2\right)^{1/3}\right)^6$.
This simplifies to: $\left(\frac{d^2 y}{d x^2}\right)^3 = \left(\left(\frac{d y}{d x}\right)^4+2\right)^2$.
The highest order derivative present is $\frac{d^2 y}{d x^2}$,so the order is $2$.
The power of the highest order derivative after rationalizing is $3$,so the degree is $3$.
Therefore,the order is $2$ and the degree is $3$.

(D) Given the differential equation: $\left(\frac{d^3 y}{d x^3}\right)^{5/4} = \left(\frac{d^2 y}{d x^2}\right)^{4/3}$.
To eliminate the fractional exponents,raise both sides to the power of $12$ (the least common multiple of $4$ and $3$):
$\left(\left(\frac{d^3 y}{d x^3}\right)^{5/4}\right)^{12} = \left(\left(\frac{d^2 y}{d x^2}\right)^{4/3}\right)^{12}$.
This simplifies to: $\left(\frac{d^3 y}{d x^3}\right)^{15} = \left(\frac{d^2 y}{d x^2}\right)^{16}$.
The highest order derivative present is $\frac{d^3 y}{d x^3}$,so the order is $3$.
The degree is the power of the highest order derivative after the equation is made free from radicals and fractions,which is $15$.
Thus,the order is $3$ and the degree is $15$.

(A) The given differential equation is $e^{\frac{d^2 y}{d x^2}} = x$.
To find the order,we identify the highest order derivative present in the equation,which is $\frac{d^2 y}{d x^2}$. Thus,the order is $2$.
To find the degree,the differential equation must be a polynomial in terms of its derivatives. Since the term $\frac{d^2 y}{d x^2}$ appears in the exponent of $e$,the equation cannot be expressed as a polynomial in its derivatives.
Therefore,the degree of this differential equation is not defined.

(C) Given the differential equation: $\{1+(\frac{dy}{dx})^2\}^{\frac{3}{2}}=\frac{d^2y}{dx^2}$.
To find the degree,we must eliminate the fractional exponent by squaring both sides:
$\{1+(\frac{dy}{dx})^2\}^3 = (\frac{d^2y}{dx^2})^2$.
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order $p = 2$.
The power of the highest order derivative after making the equation a polynomial in derivatives is $2$,so the degree $q = 2$.
Therefore,$p+q = 2+2 = 4$.