Question

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

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Answer 1

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

Explanation:

Given ${\displaystyle \frac{\text{dy}}{\text{d}x}=2(x+1)}$

⇒ dy = (2x + 2)dx

Now, Integrating above,

${\displaystyle \int \text{dy}=\int (2x+2)\text{d}x}$

⇒ y = ${\displaystyle \left(\frac{x^2}{2}\right)2+2x+\text{c}}$

⇒ y = x² + 2x + c

Area bounded by y = y(x) and x-axis is ${\displaystyle \frac{4\sqrt{8}}{3}}$

Using equation (i), at x-axis (y = 0)

x² + 2x + c = 0

x = ${\displaystyle \frac{-2\pm \sqrt{4-4\text{c}}}{2}}$

⇒ x = ${\displaystyle -1\pm \sqrt{1-\text{c}}}$

A = ${\displaystyle 2{\int }_{-1}^{-1+\sqrt{1-\text{c}}}\{-{(x+1)}^2-\text{c}+1\}\text{d}x=\frac{4\sqrt{8}}{3}}$

⇒ ${\displaystyle {[\frac{-{(x+1)}^3}{3}-\text{c}x+x]}_{-1}^{-1+\sqrt{1-\text{c}}}=\frac{2\sqrt{8}}{3}}$

⇒ ${\displaystyle \frac{-{\left(\sqrt{1-\text{c}}\right)}^3}{3}-\text{c}(-1+\sqrt{1-\text{c}})+(-1+\sqrt{1-\text{c}})-\text{c}+1=\frac{2\sqrt{8}}{3}}$

⇒ ${\displaystyle -{\left(\sqrt{1-\text{c}}\right)}^3+3\text{c}-3\text{c}\sqrt{1-\text{c}}-3+3\sqrt{1-\text{c}}-3\text{c}+3=2\sqrt{8}}$

⇒ ${\displaystyle -{\left(\sqrt{1-\text{c}}\right)}^3-3\text{c}\sqrt{1-\text{c}}+3\sqrt{1-\text{c}}=2\sqrt{8}}$

⇒ c = –1

Now using Equation (i)

y = y(x) = x² + 2x – 1

⇒ y(1) = 2