Question

If a curve $y=f\left(x\right)$ passes through the point $(1,2)$ and satisfies $x\frac{dy}{dx}+y=b{x}^4$, then for what value of $b$, ${\int }_1^{2}f\left(x\right)dx=\frac{62}{5}$

• A. $5$
• B. $\frac{62}{5}$
• C. $\frac{31}{5}$
• D. $10$

Answer 1

The correct option is D

$10$

Explanation for the correct option:

Step 1. Find the value of $b$:

Given, $x\frac{dy}{dx}+y=b{x}^4$

$\Rightarrow$ $\frac{dy}{dx}+\frac{y}{x}=b{x}^3$

Here, $\begin{array}{rcl}I.F.& =& e^{\int \frac{dx}{x}}\\ & =& x\end{array}$

$\begin{array}{rcl}\therefore yx& =& \int b{x}^{4}dx\\ & =& \frac{b{x}^5}{5}+c\end{array}$

$\Rightarrow$$y=\begin{array}{l}\frac{b{x}^4}{5}\end{array}\begin{array}{l}+\end{array}\begin{array}{l}\frac{c}{x}\end{array}$

Given that, above curve passes through $(1,2)$

$\Rightarrow$$2=\frac{b}{5}+c$ …(1)

Also, ${\int }_1^{2}f\left(x\right)dx=\frac{62}{5}$

$\Rightarrow$ ${\int }_1^2\left(\frac{b{x}^4}{5}+\frac{c}{x}\right)dx=\frac{62}{5}$

$\Rightarrow$$\frac{b}{25}\times 32+c\ln 2-\frac{b}{25}=\frac{62}{5}$ …(2)

Step 2. solve equation (1) and (2), we get

$c=0$ and $\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{10}$

Hence, Option ‘D’ is Correct.