Order and degree of differential equations Questions in English

(B) To find the degree,we first express the differential equation as a polynomial in its derivatives by eliminating fractional exponents.
Given: $x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{3}}+7 \frac{d y}{d x}+y=0$.
Multiply the entire equation by $\left(\frac{d^2 y}{d x^2}\right)^{-\frac{1}{3}}$ or simply isolate the terms with fractional powers to clear them.
Let $D = \frac{d^2 y}{d x^2}$. The equation is $x D^{1/3} + 2x^2 D^{5/3} + 7 \frac{dy}{dx} + y = 0$.
Multiply by $D^{-1/3}$: $x + 2x^2 D^{4/3} + (7 \frac{dy}{dx} + y) D^{-1/3} = 0$.
To clear the fractional powers,we raise the terms to appropriate powers. The highest power of the highest order derivative $\frac{d^2 y}{d x^2}$ after rationalizing the exponents is $5$.
Specifically,multiplying by $D^{1/3}$ gives $x D^{2/3} + 2x^2 D^2 + (7 \frac{dy}{dx} + y) D^{1/3} = 0$.
Raising to the power of $3$ to eliminate the remaining fractional powers results in a polynomial where the highest power of the highest order derivative $\frac{d^2 y}{d x^2}$ is $5$.
Thus,the degree is $5$.

(D) Given the differential equation: $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^2}$
Raising both sides to the power of $6$ to remove the radicals:
$\left(\frac{d^2 y}{d x^2}\right)^{\frac{6}{2}} = \left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^{\frac{2 \times 6}{3}}$
$\Rightarrow \left(\frac{d^2 y}{d x^2}\right)^3 = \left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^4$
The order of a differential equation is the highest derivative present,which 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. Since the term $\sin \left(\frac{d y}{d x}\right)$ is a transcendental function of the derivative,the equation cannot be expressed as a polynomial in derivatives.
Therefore,the degree is not defined.

(D) Step $1$: Analyze the Assertion $(A)$. The given differential equation is $y'' + 2xy' + \log_e\left(\frac{dy}{dx}\right) = 0$.
For a differential equation to have a defined degree,it must be a polynomial in its derivatives. The term $\log_e\left(\frac{dy}{dx}\right)$ makes the equation non-polynomial in terms of the derivative $\frac{dy}{dx}$. Therefore,the degree of this differential equation is not defined.
Thus,the Assertion $(A)$ is false.
Step $2$: Analyze the Reason $(R)$. The definition provided in the Reason states that the degree is the highest power of the highest order derivative after expressing the equation as a polynomial in differential coefficients. This is the standard mathematical definition of the degree of a differential equation.
Thus,the Reason $(R)$ is true.
Conclusion: Since $(A)$ is false and $(R)$ is true,the correct option is $(D)$.

(A) To find the number of arbitrary constants in the general solution of a differential equation,we need to determine the order of the differential equation. The number of arbitrary constants is equal to the order of the differential equation.
First,we rewrite the given equation to remove the fractional exponent:
$\left(\frac{d^4 y}{d x^4}+\frac{d^2 y}{d x^2}\right)^{3 / 2}=5 \frac{d^3 y}{d x^3}$
Squaring both sides,we get:
$\left(\frac{d^4 y}{d x^4}+\frac{d^2 y}{d x^2}\right)^3 = 25 \left(\frac{d^3 y}{d x^3}\right)^2$
The highest order derivative present in the equation is $\frac{d^4 y}{d x^4}$,which is of order $4$.
Since the order of the differential equation is $4$,the general solution will contain $4$ arbitrary constants.

(B) Given the general solution $y = a x^2 + b \log x$.
First,differentiate with respect to $x$: $y' = 2ax + \frac{b}{x}$.
Multiply by $x$: $x y' = 2ax^2 + b$.
Differentiate again with respect to $x$: $x y'' + y' = 4ax$.
From $y' = 2ax + \frac{b}{x}$,we have $b = x y' - 2ax^2$.
Substitute $a = \frac{x y' - b}{2x^2}$ into the equation,or more simply,eliminate constants $a$ and $b$.
From $y = ax^2 + b \log x$,we have $y' = 2ax + \frac{b}{x}$ and $y'' = 2a - \frac{b}{x^2}$.
Solving for $a$ and $b$: $x^2 y'' = 2ax^2 - b$ and $y = ax^2 + b \log x$.
Adding these: $x^2 y'' + y = 3ax^2 + b(\log x - 1)$.
Actually,the differential equation is $x^2 y'' + x y' - 4y = 0$.
Comparing with $f(x) y'' + g(x) y' = \frac{4y}{x}$,we get $f(x) = x^2$ and $g(x) = x$.
The order $l = 2$ and the degree $m = 1$.
Thus,$f(m) + g(m) = f(1) + g(1) = 1^2 + 1 = 2$.
Since $l = 2$,$f(m) + g(m) = l$.

(D) Given the differential equation: $\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2}=0$.
Rearranging the terms,we get: $\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2} = -\left(\frac{d^2 y}{d x^2}+y\right)$.
Squaring both sides to eliminate the fractional exponent: $\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^3 = \left(\frac{d^2 y}{d x^2}+y\right)^2$.
The highest order derivative present in the equation is $\frac{d^3 y}{d x^3}$,so the order is $3$.
The power to which the highest order derivative is raised after the equation is made free from radicals and fractions is $3$.
Therefore,the order is $3$ and the degree is $3$.

(C) For $D_1: y=4 \frac{dy}{dx}+3x \frac{dx}{dy}$. Multiplying by $\frac{dy}{dx}$,we get $y \frac{dy}{dx}=4(\frac{dy}{dx})^2+3x$. The highest derivative is $\frac{dy}{dx}$,so order is $1$. The power of the highest derivative is $2$,so degree is $2$.
For $D_2: \frac{d^2y}{dx^2}=(3+(\frac{dy}{dx})^2)^{\frac{4}{3}}$. Cubing both sides,we get $(\frac{d^2y}{dx^2})^3=(3+(\frac{dy}{dx})^2)^4$. The highest derivative is $\frac{d^2y}{dx^2}$,so order is $2$. The power of the highest derivative is $3$,so degree is $3$.
For $D_3: [1+(\frac{dy}{dx})]^2=(\frac{dy}{dx})^2$. Expanding,$1+(\frac{dy}{dx})^2+2\frac{dy}{dx}=(\frac{dy}{dx})^2$,which simplifies to $1+2\frac{dy}{dx}=0$. The highest derivative is $\frac{dy}{dx}$,so order is $1$. The power of the highest derivative is $1$,so degree is $1$.
Sum of orders $= 1+2+1 = 4$.
Sum of degrees $= 2+3+1 = 6$.
Required ratio $= \frac{4}{6} = \frac{2}{3}$.

(A) The equation of the family of all concentric circles centered at $(h, k)$ is $(x-h)^2 + (y-k)^2 = r^2$.
Since $(h, k)$ are fixed constants,the only arbitrary constant (parameter) in this equation is $r$.
The order of a differential equation is equal to the number of arbitrary constants present in the general solution of the family of curves.
Since there is only $1$ arbitrary constant $(r)$,the order of the differential equation is $1$.

(D) The given differential equation is $\left(\frac{d^2 y}{d x^2}\right)^{\frac{4}{3}}+x\left(\frac{d y}{d x}\right)^2-y \cos \left(\frac{d y}{d x}\right)=0$.
To define the degree of a differential equation,it must be a polynomial equation in terms of 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.

(C) Given,$y^{2}=2 d(x+\sqrt{d})$ $(i)$
Differentiating with respect to $x$,we get:
$2y y_{1} = 2d \Rightarrow d = y y_{1}$
Substituting $d = y y_{1}$ into equation $(i)$:
$y^{2} = 2(y y_{1})(x + \sqrt{y y_{1}})$
Rearranging the terms to isolate the radical:
$y^{2} - 2y y_{1} x = 2y y_{1} \sqrt{y y_{1}}$
Squaring both sides:
$(y^{2} - 2y y_{1} x)^{2} = (2y y_{1})^{2} (y y_{1})$
$(y^{2} - 2y y_{1} x)^{2} = 4(y y_{1})^{3}$
The highest derivative present is $y_{1}$ (order $1$),and the highest power of the highest derivative is $3$. Therefore,the degree of the differential equation is $3$.

(B) The given differential equation is $y = x(\frac{dy}{dx})^2 + \frac{dy}{dx}$.
To find the order and degree,we first express the equation in a form where the derivatives are free from radicals and fractions.
The highest order derivative present in the equation is $\frac{dy}{dx}$,which is of order $1$.
Since the highest order derivative is $\frac{dy}{dx}$,the order of the differential equation is $1$.
The degree of a differential equation is the power of the highest order derivative when the equation is written as a polynomial in derivatives.
Here,the highest power of the derivative $\frac{dy}{dx}$ is $2$.
Therefore,the degree is $2$ and the order is $1$.

(C) The given differential equation is $x = 1 + \left(\frac{dy}{dx}\right) + \frac{1}{2!} \left(\frac{dy}{dx}\right)^2 + \frac{1}{3!} \left(\frac{dy}{dx}\right)^3 + \dots$
This series is the expansion of the exponential function $e^{\frac{dy}{dx}}$.
Thus,the equation can be written as $x = e^{\frac{dy}{dx}}$.
Taking the natural logarithm on both sides,we get $\ln(x) = \frac{dy}{dx}$.
The differential equation is $\frac{dy}{dx} = \ln(x)$.
In this equation,the highest order derivative is $\frac{dy}{dx}$,which is of order $1$.
The power of the highest order derivative is $1$.
Therefore,the degree of the differential equation is $1$.

(B) The given differential equation is $\frac{d^2 y}{d x^2} = \sqrt{1 - (\frac{dy}{dx})^2}$.
To find the order,we identify the highest order derivative present in the equation.
The highest order derivative is $\frac{d^2 y}{d x^2}$,which represents the second derivative of $y$ with respect to $x$.
The order of a differential equation is defined as the order of the highest derivative involved in the equation.
Since the highest derivative is of order $2$,the order of the given differential equation is $2$.

(D) To find the degree,we must first express the differential equation as a polynomial in terms of its derivatives.
Given equation: $(1 + \frac{dy}{dx})^2 = (\frac{d^3y}{dx^3})^{1/3}$.
Raise both sides to the power of $3$ to eliminate the fractional exponent: $((1 + \frac{dy}{dx})^2)^3 = ((\frac{d^3y}{dx^3})^{1/3})^3$.
This simplifies to: $(1 + \frac{dy}{dx})^6 = \frac{d^3y}{dx^3}$.
The highest order derivative present is $\frac{d^3y}{dx^3}$,which is of order $3$.
The degree of a differential equation is the power of the highest order derivative when the equation is written as a polynomial in its derivatives.
Here,the power of $\frac{d^3y}{dx^3}$ is $1$. Therefore,the degree is $1$.

(D) To find the order and degree,we must first eliminate the radicals by raising both sides to the power of $6$ (the least common multiple of $2$ and $3$).
Given equation: $(1 + (y'')^2)^{1/2} = (x + (y')^3)^{1/3}$.
Raising both sides to the power of $6$:
$((1 + (y'')^2)^{1/2})^6 = ((x + (y')^3)^{1/3})^6$
$(1 + (y'')^2)^3 = (x + (y')^3)^2$.
Expanding the left side: $1 + 3(y'')^2 + 3(y'')^4 + (y'')^6 = (x + (y')^3)^2$.
The highest order derivative present is $y'' = \frac{d^2y}{dx^2}$,so the order is $2$.
The exponent of the highest order derivative after rationalizing the equation is $6$. Therefore,the degree is $6$.

(D) The 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 $0$ arbitrary constants.