JEE PYQ: Differential Equation Question 32
Question 32 - 2020 (09 Jan Shift 1)
If for $x \ge 0$, $y = y(x)$ is the solution of the differential equation $(x+1),dy = ((x+1)^2 + y - 3),dx$, $y(2) = 0$, then $y(3)$ is equal to ______.
Show Answer
Answer: 3
Solution
$\frac{dy}{dx} = (x+1) + \frac{y-3}{x+1}$. Let $Y = y - 3$: $\frac{dY}{dx} = x + 1 + \frac{Y}{x+1}$. I.F. $= \frac{1}{x+1}$. Solution: $\frac{Y}{x+1} = x + 3(x+1)^{-1} + C$. At $(2,0)$: $Y = -3$, so $\frac{-3}{3} = 2 + 1 + C$, $C = -4$… Wait, let me use the solution from the image: $\frac{y}{1+x} = x + \frac{3}{(1+x)} + C$. At $(2,0)$: $0 = 2 + 1 + C$, $C = -3$. At $x = 3$: $\frac{y}{4} = 3 + \frac{3}{4} - 3 = \frac{3}{4}$, $y = 3$.
BETA
