Let $y=y(x)$ be the solution curve of the differential equation $$ \frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$$, which passes through the point $(0,1)$. Then $y(1)$ is equal to :
Solution
$\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$,
<br/><br/>Integrating factor I.F. $=e^{\int \frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6} d x}$
<br/><br/>$$
\begin{aligned}
& \text { Let } \frac{2 x^{2}+11 x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3} \\\\
& A=2, B=1, C=-1 \\\\
& \text { I.F. }=e^{(2 \ln |x+1|+\ln |x+2|-\ln |x+3|)} \\\\
& =\frac{(x+1)^{2}(x+2)}{x+3}
\end{aligned}
$$
<br/><br/>Solution of differential equation
<br/><br/>$$
\begin{aligned}
&y \cdot \frac{(x+1)^{2}(x+2)}{x+3}=\int(x+1)(x+2) d x \\\\
&y \frac{(x+1)^{2}(x+2)}{x+3}=\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+2 x+c
\end{aligned}
$$
<br/><br/>Curve passes through $(0,1)$
<br/><br/>$$
\begin{gathered}
1 \times \frac{1 \times 2}{3}=0+c \Rightarrow c=\frac{2}{3} \\\\
\text { So, } y(1)=\frac{\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}}{\frac{\left(2^{2} \times 3\right)}{4}}=\frac{3}{2}
\end{gathered}
$$
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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