JEE Main 2022 · 29 Jul · Shift 2 · Mathematics Q10

← JEE Main 2022 · 29 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 10 of 90

Let $y=y(x)$ be the solution curve of the differential equation $\dfrac{dy}{dx}+\left(\dfrac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\dfrac{(x+3)}{x+1}$ , x > −1, which passes through the point (0, 1). Then y(1) is equal to :

\dfrac12

\dfrac32

\dfrac52

\dfrac72

Answer and worked solution

Official answer: B

Partial fractions. $\dfrac{2x^2+11x+13}{(x+1)(x+2)(x+3)}=\dfrac2{x+1}+\dfrac1{x+2}-\dfrac1{x+3}$ , so the integrating factor is $\dfrac{(x+1)^2(x+2)}{x+3}$ .

Integrate. $y\cdot\dfrac{(x+1)^2(x+2)}{x+3}=\displaystyle\int(x+1)(x+2)dx=\frac{x^3}3+\frac{3x^2}2+2x+C$ ; (0, 1) gives $C=\frac23$ .

At x = 1: $3y=\frac13+\frac32+2+\frac23=\frac92$ , so $y=\frac32$ .

(B)

Topics:Linear first-order equations Partial fractions

← Previous question Next question →

JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 10