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The given differential equation is: \[ (x + 2y^3)\frac{dy}{dx} = 2y \] We rearrange the equation to the linear form $\frac{dx}{dy} + P(y)x = Q(y)$: \[ \frac{dx}{dy} = \frac{x + 2y^3}{2y} = \frac{x}{2y} + \frac{2y^3}{2y} = \frac{1}{2y}x + y^2 \] \[ \frac{dx}{dy} - \frac{1}{2y}x = y^2 \] This is a linear differential equation in $x$ with $P(y) = -\frac{1}{2y}$. The integrating factor (IF) is given by: \[ IF = e^{\int P(y) dy} \] Substitute $P(y)$: \[ IF = e^{\int \left(-\frac{1}{2y}\right) dy} = e^{-\frac{1}{2} \int \frac{1}{y} dy} \] \[ IF = e^{-\frac{1}{2} \ln |y|} \] Using the logarithm property $a \ln b = \ln b^a$: \[ IF = e^{\ln |y|^{-\frac{1}{2}}} = |y|^{-\frac{1}{2}} \] Assuming $y > 0$, the integrating factor is: \[ IF = \frac{1}{\sqrt{y}} \]

The given differential equation is:

$\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)^3 = 0$

Step 1: Expand the derivative

$\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)^3 = 3\left(\frac{dy}{dx}\right)^2 \frac{d^2y}{dx^2} = 0$

Step 2: Identify order and degree

• The highest order derivative is $\dfrac{d^2y}{dx^2}$, so the order is $p = 2$.
• The equation is a polynomial in derivatives, and the degree of the highest derivative is $1$, so $q = 1$.

Step 3: Compute $(p - q)$

$(p - q) = 2 - 1 = 1$

∴ Final Answer: $(p - q) = 1$

The order of a differential equation is the order of the highest derivative present in the equation.The degree of a differential equation is the power of the highest order derivative, provided the differential equation is a polynomial equation in its derivatives.

For the degree to be defined, the equation must be free from terms like $\sin(\frac{dy}{dx})$, $e^{\frac{dy}{dx}}$, $\log(\frac{d^2y}{dx^2})$, etc. (i.e., the derivatives must not appear inside transcendental functions).In the given equation, the term $x\sin\left(\frac{dy}{dx}\right)$ involves the first derivative $\frac{dy}{dx}$ inside the transcendental function $\sin()$.Since the equation is not a polynomial in its derivatives, the Degree is not defined

\(\int_{0}^{4} \sqrt{y} dy\)

The area (\(A\)) of the region enclosed by the curve \(y = f(x)\), the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\) is given by the definite integral:\[A = \int_{a}^{b} f(x) dx\] \[A = \int_{0}^{4} x^{1/2} dx\] \[A = \frac{2}{3} \left[ x^{3/2} \right]_{0}^{4}\] \[A = \frac{2}{3} \left[ (4)^{3/2} - (0)^{3/2} \right]\] \[A = \frac{16}{3}\]

The region is bounded by the parabola $y^2=4x$, the X-axis ($y=0$), $x=0$, and $x=1$. Since the parabola is symmetric about the X-axis, the total area can be found by calculating the area of the upper half (where $y = 2\sqrt{x}$) and multiplying by 2. 1. From $y^2=4x$, we get $y = \pm \sqrt{4x} = \pm 2\sqrt{x}$. 2. The area $A$ is given by the integral of $y$ with respect to $x$ from $x=0$ to $x=1$. Due to symmetry, we calculate the area for $y=2\sqrt{x}$ and multiply by 2. $$A = 2 \int_{0}^{1} 2\sqrt{x} \, dx$$ 3. Evaluate the integral: $$A = 4 \int_{0}^{1} x^{1/2} \, dx$$ $$A = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1}$$ $$A = 4 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1}$$ 4. Apply the limits of integration: $$A = \frac{8}{3} \left( (1)^{3/2} - (0)^{3/2} \right)$$ $$A = \frac{8}{3} (1 - 0)$$ $$A = \frac{8}{3}$$ The final answer is $\boxed{\frac{8}{3}}$.

The curve \( y^2 = x \) is symmetric about the x-axis. The shaded area is the area above the x-axis: Answer:
\( \int_0^4 \sqrt{x} \, dx \)

**Correct Option if MCQ:** B **Reasoning:** * \(y = \sqrt{x}\) * Area = \(\int_{0}^{1} \sqrt{x} dx\) * Area = \([\frac{2}{3}x^{\frac{3}{2}}]_{0}^{1} = \frac{2}{3}\)