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

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

$\frac{5}{2}$

$\frac{7}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} \frac{dy}{dx} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1} \\ I . F . = e^{\int p (x) dx} \\ \int p (x) dx = \int \frac{(2 x^{2} + 11 x + 13) dx}{(x + 1) (x + 2) (x + 3)} \end{array}$
Using partial fraction
$\begin{array}{l} \frac{2 x^{2} + 11 x + 13}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3} \\ A = \frac{4}{2} = 2 \\ B = 1 \\ C = - 1 \\ ∵ \int p (x) dx = Aln (x + 1) + Bln (x + 2) + cln (x + 3) \\ = (\frac{(x + 1)^{2} (x + 2)}{x + 3}) \end{array}$
I.F $= e^{\int p (x) dx} = \frac{(x + 1)^{2} (x + 2)}{(x + 3)}$
Solution $y (IF) = \int Q \cdot (IF) dx$
$\begin{array}{l} y (\frac{(x + 1)^{2} (x + 2)}{x + 3}) = \int (\frac{x + 3}{x + 1}) \frac{(x + 1)^{2} (x + 2)}{(x + 3)} dx \\ \Rightarrow y (\frac{(x + 1)^{2} (x + 2)}{x + 3}) = \frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x + c \end{array}$
Passes through (0, 1) $C = \frac{2}{3}$
Now put x=1
$\Rightarrow y (1) = \frac{3}{2}$