JEE Main 2025 — Differential Equations Question with Solution
JEE Main 2025 (24 Jan Shift 1)
Question
Let be a differentiable function such that . Then is equal to ______.
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Show full solutionCorrect answer: 19
Correct answer
19
Step-by-step explanation
Differentiating both side
$\begin{aligned}
& 4(x+2) f(x)+2 f(x)(x+2)^2-6(x+2)=10(x+2) f(x) \\
& =(x+2) \frac{d y}{d x}-3 y=3 \\
& \frac{1}{3} \int \frac{d y}{y+1}=\int \frac{d x}{x+2} \\
& \ln |y+1=3 \ln | x+2 \mid+\ln c \\
& y+1=(x+2)^3 c \\
& \because y(0)=\frac{3}{2} \\
& \Rightarrow \frac{5}{16}=c \\
& \therefore \quad y=\frac{5}{16}(x+2)^3-1 \\
& y(2)=\frac{5}{16} \times 64-1=19
\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Differential Equations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.