Let f be a differentiable function such that 2(x+2)^2 f(x)-3(x+2)^2=10 ∫_0^x(t+2) f(t) d t, x ≥ 0 . Then f(2) is equal to ______.

Correct answer: 0 2(x+2)^2 f(x)-3(x+2)^2=10 ∫_0^x(t+2) f(t) d t Differentiating both side aligned & 4(x+2) f(x)+2 f(x)(x+2)^2-6(x+2)=10(x+2) f(x) & =(x+2) (d y)/(d x)-3 y=3 & (1)/(3) ∫ (d y)/(y+1)=∫ (d x)/(x+2) & |y+1=3 | x+2 + c & y+1=(x+2)^3 c & y(0)=(3)/(2) & (5)/(16)=c & y=(5)/(16)(x+2)^3-1 & y(2)=(5)/(16) × 64-1=19 aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Differential Equations

Let f be a differentiable function such that 2(x+2)^2...

Let

f

be a differentiable function such that

2(x+2)^2 f(x)-3(x+2)^2=10 \int_0^x(t+2) f(t) d t, x \geq 0

. Then

f(2)

is equal to ______.

Correct answer

Step-by-step solution

2(x+2)^2 f(x)-3(x+2)^2=10 \int_0^x(t+2) f(t) d t

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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