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

The Correct Option is B

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