Question

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

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

Answer: (B)

$\frac{3}{2}$

Answer 2

$\frac{dy}{dx} + \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6}y = (x+3)(x+1), \quad x > -1.$
Integrating factor I.F

$e^{\int \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6} \, dx}$

Let $\frac{22 + 11x + 13}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}$
$A = 2, B = 1, C = –1$

$I.F. = e^{(2\ln|x+1|+\ln|x+2|-\ln|x+3|)}$

$\frac{(x+1)^2 \cdot (x+2)}{x+3}$
Solution of differential equation

$y \cdot \frac{(x+1)^2(x+2)}{x+3} = \int (x+1)(x+2) \, dx$

$y \cdot \frac{(x+1)^2(x+2)}{x+3} = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + c$
Curve passes through (0, 1)

$1 \times 1 \times \frac{2}{3} = 0 + c \Rightarrow c = \frac{2}{3}$

So, $y(1) = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} \div \frac{2^2 \times 3}{4}$
$=\frac{3}{2}$
So, the correct option is (B): $\frac{3}{2}$