Let y = y(x) be the solution curve of the differential equation...

Given,

$\dfrac{dy}{dx} + \left(\dfrac{2{x}^{2} + 11x + 13}{{x}^{3} + 6{x}^{2} + 11x + 6}\right)y = \dfrac{x + 3}{x + 1}$

Now comparing with $\dfrac{dy}{dx} + y \times p\left(x\right) = q\left(x\right)$
We get $p\left(x\right) = \dfrac{2{x}^{2} + 11x + 13}{{x}^{3} + 6{x}^{2} + 11x + 6}$

So $IF = {e}^{∫ \dfrac{2{x}^{2} + 11x + 13}{{x}^{3} + 6{x}^{2} + 11x + 6}dx}$

Now finding the integration we get,

$∫ p\left(x\right)dx = ∫ \dfrac{\left(2{x}^{2} + 11x + 13\right)dx}{\left(x + 1\right)\left(x + 2\right)\left(x + 3\right)}$

Using partial fraction we get,

$\dfrac{2{x}^{2} + 11x + 13}{\left(x + 1\right)\left(x + 2\right)\left(x + 3\right)} = \dfrac{A}{x + 1} + \dfrac{B}{x + 2} + \dfrac{C}{x + 3}$

On solving we get A = $\dfrac{4}{2} =$2$,$B = 1 and C = - 1$</p><p>So,$∫ p\left(x\right)$dx = Aln\left(x + 1\right)$ + Bln\left(x + 2\right)$+ cln\left(x + 3\right)$

= \ln \left(\dfrac{{\left(x + 1\right)}^{2}\left(x + 2\right)}{x + 3}\right)$</p><p>So,$IF = {e}^{∫ p\left(x\right)dx} = \dfrac{{\left(x + 1\right)}^{2}\left(x + 2\right)}{\left(x + 3\right)}$</p><p>Now solution of differential equation is given by</p><p>$y \times IF = ∫ Q \times IFdx$</p><p>$⇒ y \times \left(\dfrac{{\left(x + 1\right)}^{2}\left(x + 2\right)}{x + 3}\right) = ∫ \left(\dfrac{x + 3}{x + 1}\right)\dfrac{{\left(x + 1\right)}^{2}\left(x + 2\right)}{\left(x + 3\right)}dx$</p><p>$⇒ y \times \left(\dfrac{{\left(x + 1\right)}^{2}\left(x + 2\right)}{x + 3}\right) = \dfrac{{x}^{3}}{3} + \dfrac{3{x}^{2}}{2} + 2x + c$</p><p>Now given curve passes through \left(0, 1\right)$,

So, $⇒ 1 \times \left(\dfrac{{\left(0 + 1\right)}^{2}\left(0 + 2\right)}{0 + 3}\right) = \dfrac{{0}^{3}}{3} + \dfrac{3 \times {0}^{2}}{2} + 2 \times 0 + c ⇒ c = \dfrac{2}{3}$

So, curve becomes $y \times \left(\dfrac{{\left(x + 1\right)}^{2}\left(x + 2\right)}{x + 3}\right) = \dfrac{{x}^{3}}{3} + \dfrac{3{x}^{2}}{2} + 2x + \dfrac{2}{3}$

Now put $x = 1$ we get, $y \times \left(\dfrac{{\left(1 + 1\right)}^{2}\left(1 + 2\right)}{1 + 3}\right) = \dfrac{{1}^{3}}{3} + \dfrac{3 \times {1}^{2}}{2} + 2 \times 1 + \dfrac{2}{3}$

$⇒ y \times 3 = 1 + \dfrac{3}{2} + 2 ⇒ y \times 3 = \dfrac{9}{2}$

$⇒ y\left(1\right) = \dfrac{3}{2}$