Question

If \( y = f(x) \) passes through \( (1, 2) \) and \( x \frac{dy}{dx} + y = b x^4 \), then for what value of \( b \), \( \displaystyle \int_{1}^{2} f(x)\,dx = \frac{62}{5} \) ?

• A. 10
• B. 31/5
• C. 5
• D. 62/5

Hint:

Recognizing the LHS as a perfect derivative ($d(xy)$) makes this first-order differential equation trivial to solve without calculating an integrating factor.

Answer 1

The Correct Option is A

Step 1: $x \frac{dy}{dx} + y = b x^4$ is $\frac{d}{dx}(xy) = bx^4$.
Step 2: Integrate: $xy = \frac{bx^5}{5} + C$.
Step 3: Pass through $(1, 2)$: $1(2) = \frac{b}{5} + C \Rightarrow C = 2 - \frac{b}{5}$.
Step 4: $f(x) = \frac{bx^4}{5} + \frac{2 - b/5}{x}$.
Step 5: $\int_1^2 [\frac{bx^4}{5} + \frac{2 - b/5}{x}] dx = [\frac{bx^5}{25} + (2 - \frac{b}{5})\ln x]_1^2 = \frac{31b}{25} + (2 - \frac{b}{5})\ln 2$.
Step 6: Given integral $= 62/5$. For this to be independent of $\ln 2$, $2 - b/5 = 0 \Rightarrow b = 10$.
Step 7: Check: $\frac{31(10)}{25} = \frac{310}{25} = \frac{62}{5}$. Correct.