For each of the following differential equations, determine its order, degree (if it exists):

(i) $\dfrac{dy}{dx}+xy=\cot x$

(ii) $\left(\dfrac{d^3y}{dx^3}\right)^{2/3}-3\dfrac{d^2y}{dx^2}+5\dfrac{dy}{dx}+4=0$

(iii) $\left(\dfrac{d^2y}{dx^2}\right)^2+\left(\dfrac{dy}{dx}\right

Question

For each of the following differential equations, determine its order, degree (if it exists):

(i) $\dfrac{dy}{dx}+xy=\cot x$

(ii) $\left(\dfrac{d^3y}{dx^3}\right)^{2/3}-3\dfrac{d^2y}{dx^2}+5\dfrac{dy}{dx}+4=0$

(iii) $\left(\dfrac{d^2y}{dx^2}\right)^2+\left(\dfrac{dy}{dx}\right)^2=x\sin\!\left(\dfrac{d^2y}{dx^2}\right)$

(iv) $\sqrt{\dfrac{dy}{dx}}-4\dfrac{dy}{dx}-7x=0$

(v) $y\left(\dfrac{dy}{dx}\right)=\dfrac{x}{\left(\dfrac{dy}{dx}\right)+\left(\dfrac{dy}{dx}\right)^3}$

(vi) $x^2\dfrac{d^2y}{dx^2}+\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{1/2}=0$

(vii) $\left(\dfrac{d^2y}{dx^2}\right)^3=\sqrt{1+\dfrac{dy}{dx}}$

(viii) $\dfrac{d^2y}{dx^2}=xy+\cos\!\left(\dfrac{dy}{dx}\right)$

(ix) $\dfrac{d^2y}{dx^2}+5\dfrac{dy}{dx}+\displaystyle\int y\,dx=x^3$

(x) $x=e^{xy\left(\frac{dy}{dx}\right)}$

Accepted Answer

> [!TLDR]
>
> Work each part by (a) identifying the highest-order derivative, (b) clearing any radical/fractional power on a derivative by raising both sides to a suitable power, and (c) checking whether any derivative (highest **or** lower order) is trapped inside a transcendental function — if so the degree is not defined.

**Step 1. (i) $\dfrac{dy}{dx}+xy=\cot x$.** Already polynomial; highest derivative $\dfrac{dy}{dx}$ has power $1$. Order $1$, degree $1$.

**Step 2. (ii) $\left(\dfrac{d^3y}{dx^3}\right)^{2/3}-3\dfrac{d^2y}{dx^2}+5\dfrac{dy}{dx}+4=0$.** Isolate the fractional-power term and cube: $\left(\dfrac{d^3y}{dx^3}\right)^2=\left(3\dfrac{d^2y}{dx^2}-5\dfrac{dy}{dx}-4\right)^3$. Highest derivative is now $\dfrac{d^3y}{dx^3}$ (order $3$), to power $2$: degree $2$.

**Step 3. (iii) $\left(\dfrac{d^2y}{dx^2}\right)^2+\left(\dfrac{dy}{dx}\right)^2=x\sin\!\left(\dfrac{d^2y}{dx^2}\right)$.** The highest-order derivative $\dfrac{d^2y}{dx^2}$ appears as the argument of $\sin$, so the equation cannot be reduced to polynomial form. Order $2$, degree not defined.

**Step 4. (iv) $\sqrt{\dfrac{dy}{dx}}-4\dfrac{dy}{dx}-7x=0$.** Isolate the root: $\sqrt{\dfrac{dy}{dx}}=4\dfrac{dy}{dx}+7x$; square both sides: $\dfrac{dy}{dx}=\left(4\dfrac{dy}{dx}+7x\right)^2$. This is polynomial, with $\dfrac{dy}{dx}$ (order $1$) appearing to the highest power $2$ (from expanding the square). Order $1$, degree $2$.

**Step 5. (v) $y\dfrac{dy}{dx}=\dfrac{x}{\dfrac{dy}{dx}+\left(\dfrac{dy}{dx}\right)^3}$.** Cross-multiply: $y\dfrac{dy}{dx}\left[\dfrac{dy}{dx}+\left(\dfrac{dy}{dx}\right)^3\right]=x\ \Longrightarrow\ y\left(\dfrac{dy}{dx}\right)^2+y\left(\dfrac{dy}{dx}\right)^4=x$. Order $1$; the highest power of $\dfrac{dy}{dx}$ is $4$: degree $4$.

**Step 6. (vi) $x^2\dfrac{d^2y}{dx^2}+\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{1/2}=0$.** Isolate and square: $\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{1/2}=-x^2\dfrac{d^2y}{dx^2}\ \Longrightarrow\ 1+\left(\dfrac{dy}{dx}\right)^2=x^4\

Q.For each of the following differential equations, determine its order, degree (if it exists):

Concept understanding — Order and Degree of a Differential Equation

Q.For each of the following differential equations, determine its order, degree (if it exists):

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