JEE PYQ: Differential Equation Question 5
Question 5 - 2021 (17 Mar Shift 2)
If the curve $y = y(x)$ is the solution of the differential equation $2(x^2 + x^{5/4}),dy - y(x + x^{1/4}),dx = 2x^{9/4},dx$, $x > 0$, which passes through the point $\left(1, 1 - \frac{4}{3}\log_e 2\right)$, then the value of $y(16)$ is equal to:
(1) $4\left(\frac{31}{3} + \frac{8}{3}\log_e 3\right)$
(2) $\left(\frac{31}{3} + \frac{8}{3}\log_e 3\right)$
(3) $4\left(\frac{31}{3} - \frac{8}{3}\log_e 3\right)$
(4) $\left(\frac{31}{3} - \frac{8}{3}\log_e 3\right)$
Show Answer
Answer: (3)
Solution
Rewrite as $\frac{dy}{dx} - \frac{y}{2x} = \frac{x^{9/4}}{x^{5/4}(x^{3/4}+1)}$. I.F. $= x^{-1/2}$. Solving with substitution $x = t^4$: $yx^{-1/2} = \frac{4}{3}x^{5/4} - \frac{4}{3}\ln(x^{3/4}+1) + C$. Using initial condition: $C = -\frac{1}{3}$. At $x = 16$: $y(16) = 4\left(\frac{31}{3} - \frac{8}{3}\log_e 3\right)$.
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