Question

Let the curve $y=y\left(x\right)$ be the solution of the differential equation,$\frac{dy}{dx}=2(x+1)$. If the numerical value of area bounded by the curve $y=y\left(x\right)$ and the $x$-axis is $\frac{4\sqrt{8}}{3}$, then the value of $y\left(1\right)$ is equal to ___

Answer 1

We need to solve the given differential equation and find the value of $y\left(1\right)$:

Step 1: Given information:

Illustrating the given information

The given curve $y=y\left(x\right)$ and the given differential equation $\frac{dy}{dx}=2(x+1)$

Step 2: Calculating the value of $\mathrm{x}$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=2\left(x+1\right) \\ \Rightarrow \int dy=\int 2\left(x+1\right)dx \\ \Rightarrow y\left(x\right)=x^2+2x+C..,...\left(i\right) \\ \Rightarrow y\left(x\right)=\left(x^2+2x+1^2\right)-1^2+C \\ \Rightarrow y\left(x\right)=-{\left(x+1\right)}^2-C+1...\left(ii\right) \end{gathered}$$

If $x^2+2x+C=0$

$$\begin{gathered} \Rightarrow x=\frac{-2\pm \sqrt{2^2-4C}}{2\times 1} \\ \Rightarrow x=-1+\sqrt{1-C} \end{gathered}$$

Step 3: Calculating the area by finite integration

The given area is$\frac{4\sqrt{8}}{3}$

Therefore, from the figure

$$\begin{gathered} \Rightarrow 2{\int }_{-1}^{-1+\sqrt{1-C}}y\left(x\right)dx=\frac{4\sqrt{8}}{3} \\ \Rightarrow 2{\int }_{-1}^{-1+\sqrt{1-C}}\left[-{\left(x+1\right)}^2-C+1\right]dx=\frac{4\sqrt{8}}{3} \\ \Rightarrow 2{\left[-\frac{{\left(x+1\right)}^3}{3}-Cx+x\right]}_{-1}^{-1+\sqrt{1-C}}=\frac{4\sqrt{8}}{3} \\ \Rightarrow -{\left(\sqrt{1-C}\right)}^3+3C-3C\sqrt{1-C}-3+3\sqrt{1-C}-3C+3=2\sqrt{8} \\ \Rightarrow C=-1 \\ \Rightarrow y\left(x\right)=x^2+2x-1 \\ \Rightarrow y\left(1\right)=2^2+2\times 1-1[x=1] \\ \Rightarrow y\left(1\right)=2 \end{gathered}$$

Hence, the value of $y\left(1\right)$ is equal to $2$.