If a differential equation is given by `dy/dx=2(x+1)` and area bounded by `y(x)` with x-axis is `(4*sqrt8)/3` then `y(1)` is

Answer

To solve the differential equation \( \frac{dy}{dx} = 2(x+1) \) and find \( y(1) \) given that the area bounded by \( y(x) \) with the x-axis is \( \frac{4\sqrt{8}}{3} \), we will follow these steps:

### Step 1: Integrate the Differential Equation
We start with the differential equation:

\[ \frac{dy}{dx} = 2(x + 1) \]

We can separate variables and integrate both sides:

\[ dy = 2(x + 1)dx \]

Integrating both sides gives:

\[ y = \int 2(x + 1)dx = 2\left(\frac{x^2}{2} + x\right) + C = x^2 + 2x + C \]

### Step 2: Determine the Area Under the Curve
The area \( A \) bounded by the curve \( y(x) \) and the x-axis can be computed using the formula for the area under the curve:

\[ A = \int_{a}^{b} y \, dx \]

Given that the area is \( \frac{4\sqrt{8}}{3} \), we need to find the limits of integration \( a \) and \( b \). Assuming the roots of the quadratic \( y = x^2 + 2x + C \) are \( \alpha \) and \( \beta \), the area can be expressed as:

\[ A = \frac{2}{3} \cdot \text{Area of Rectangle} \]

The area of the rectangle formed by the roots is given by:

\[ \text{Area of Rectangle} = \text{height} \times \text{width} = (1 - C) \cdot (\beta - \alpha) \]

### Step 3: Find the Roots of the Quadratic
The roots of the quadratic equation \( x^2 + 2x + C = 0 \) can be found using the quadratic formula:

\[ \alpha, \beta = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{4 - 4C}}{2} \]

Where \( D = 4 - 4C \) is the discriminant.

### Step 4: Calculate the Area
The width \( \beta - \alpha \) can be expressed as:

\[ \beta - \alpha = \frac{\sqrt{4 - 4C}}{1} = \sqrt{4(1 - C)} = 2\sqrt{1 - C} \]

Now substituting this into the area formula:

\[ A = \frac{2}{3} \cdot (1 - C) \cdot 2\sqrt{1 - C} = \frac{4}{3} (1 - C) \sqrt{1 - C} \]

Setting this equal to the given area \( \frac{4\sqrt{8}}{3} \):

\[ \frac{4}{3} (1 - C) \sqrt{1 - C} = \frac{4\sqrt{8}}{3} \]

### Step 5: Simplify and Solve for \( C \)
Cancelling \( \frac{4}{3} \) from both sides gives:

\[ (1 - C) \sqrt{1 - C} = \sqrt{8} \]

Let \( u = 1 - C \), then:

\[ u \sqrt{u} = \sqrt{8} \]

Squaring both sides:

\[ u^3 = 8 \implies u = 2 \]

Thus:

\[ 1 - C = 2 \implies C = -1 \]

### Step 6: Find \( y(1) \)
Now substituting \( C \) back into the equation for \( y \):

\[ y = x^2 + 2x - 1 \]

To find \( y(1) \):

\[ y(1) = 1^2 + 2(1) - 1 = 1 + 2 - 1 = 2 \]

### Final Answer
Thus, \( y(1) = 2 \).