JEE PYQ: Differential Equation Question 10
Question 10 - 2021 (24 Feb Shift 2)
If a curve $y = f(x)$ passes through the point $(1, 2)$ and satisfies $x\frac{dy}{dx} + y = bx^4$, then for what value of $b$, $\int_1^2 f(x),dx = \frac{62}{5}$?
(1) 5
(2) $\frac{62}{5}$
(3) $\frac{31}{5}$
(4) 10
Show Answer
Answer: (4)
Solution
$\frac{dy}{dx} + \frac{y}{x} = bx^3$. I.F. $= x$. Solution: $yx = \int bx^4,dx = \frac{bx^5}{5} + c$. Through $(1,2)$: $2 = \frac{b}{5} + c$. Also $\int_1^2 f(x),dx = \int_1^2 \left(\frac{bx^4}{5} + \frac{c}{x}\right),dx = \frac{b}{5}\cdot\frac{31}{5} + c\ln 2 = \frac{62}{5}$. From $c = 0$ and $b = 10$: $\frac{10 \times 31}{25} + 0 = \frac{62}{5}$. So $b = 10$.
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