Question

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

Answer 1

$\frac{dy}{dx}+\frac{(2x^2+11x+13)}{(x+1)(x+2)(x+3)}y=\frac{x+3}{x+1}$

$I.F=e^{\int \frac{2x^2+11x+13}{(x+1)(x+2)(x+3)}dx}$

$=e^{\int (\frac{2}{x+1}+\frac{1}{x+2}-\frac{1}{x+3})dx}$

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

$=e^{\ln \frac{(x+1{)}^2(x+2)}{(x+3)}}$

$I.F=\frac{(x+1{)}^2(x+2)}{(x+3)}$

So, curve

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

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

$\Rightarrow y\frac{(x+1{)}^2(x+2)}{(x+3)}=\frac{x^3}{3}+\frac{3x^2}{2}+2x+c$

$\because$ Passes through $(0,1)\Rightarrow c=\frac{2}{3}$

So solution curve is

$y\frac{(x+1{)}^2(x+2)}{(x+3)}=\frac{x^3}{3}+\frac{3x^2}{2}+2x+\frac{2}{3}$

for $x=1\Rightarrow y=\frac{3}{2}$