Find the particular solution satisfying the given condition $\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)\dfrac{dy}{dx}=2{{x}^{2}}+x$ ; $y=1$when $x=0$. \[\]

Hint: We separate the variables $x,y$ to different sides and factorize the denominator ${{x}^{3}}+{{x}^{2}}+x+1$ use partial fraction method to find $A,B,C$ in the equation expression of right hand side $\dfrac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{{{x}^{2}}+1}.$We integrate in respective sides using standard integration $\int{\dfrac{dx}{x+a}}=\log \left( x+a \right),\int{\dfrac{dx}{{{x}^{2}}+{{a}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)$ and method of substitution. \[\]

Complete step by step answer:
We are given a differential equation in the question as;
\[\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)\dfrac{dy}{dx}=2{{x}^{2}}+x\]
We are also given an initial condition $y=1$when$x=0$. Let us separate the variables in the differential equation. We have;
\[dy=\dfrac{2{{x}^{2}}+x}{{{x}^{3}}+{{x}^{2}}+x+1}dx\]
We see that we cannot integrate directly. So we have to first factorize the denominator$p\left( x \right)={{x}^{3}}+{{x}^{2}}+x+1$. So let us take ${{x}^{2}}$ common form first two terms of $p\left( x \right)$ and have;
\[\begin{align}
  & \Rightarrow dy=\dfrac{2{{x}^{2}}+x}{{{x}^{2}}\left( x+1 \right)+1\left( x+1 \right)}dx \\
 & \Rightarrow dy=\dfrac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}dx.......\left( 1 \right) \\
\end{align}\]
Now we need to use a partial fraction method to separate the polynomials in the denominator. Let us assume for some real constants $A,B,C$
\[\dfrac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{{{x}^{2}}+1}.......\left( 2 \right)\]
 We multiply both side of the above equation by $\left( x+1 \right)\left( {{x}^{2}}+1 \right)$ to have;
\[2{{x}^{2}}+x=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+1 \right)......\left( 3 \right)\]
Let us put $x=-1$ in above equation (3) and have;
\[\begin{align}
  & 2{{\left( -1 \right)}^{2}}+\left( -1 \right)=A\left\{ {{\left( -1 \right)}^{2}}+1 \right\}+0 \\
 & \Rightarrow 1=2A \\
 & \Rightarrow A=\dfrac{1}{2} \\
\end{align}\]
Let us put $x=0$ in above equation (3) and have;

\[\begin{align}
  & 2{{\left( 0 \right)}^{2}}+0=A\left( 0+1 \right)+\left( B\cdot 0+C \right)\left( 0+1 \right) \\
 & \Rightarrow 0=A+C \\
 & \Rightarrow C=-A \\
 & \Rightarrow C=-\dfrac{1}{2}\left( \because A=\dfrac{1}{2} \right) \\
\end{align}\]
Let us put $x=1$ in above equation (3) and then put $A=\dfrac{1}{2},C=\dfrac{-1}{2}$ to have;
 \[\begin{align}
  & 2{{\left( 1 \right)}^{2}}+1=A\left( {{1}^{2}}+1 \right)+\left( B\cdot 1+C \right)\left( 1+1 \right) \\
 & \Rightarrow 3=2A+2B+2C \\
 & \Rightarrow A+B+C=\dfrac{3}{2} \\
 & \Rightarrow \dfrac{1}{2}+B-\dfrac{1}{2}=\dfrac{3}{2} \\
 & \Rightarrow B=\dfrac{3}{2} \\
\end{align}\]
Let us put the obtained values of $A,B,C$ in equation (1) to have;
\[\begin{align}
  & \dfrac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{{{x}^{2}}+1} \\
 & \Rightarrow \dfrac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\dfrac{\dfrac{1}{2}}{x+1}+\dfrac{\dfrac{3}{2}x-\dfrac{1}{2}}{{{x}^{2}}+1} \\
 & \Rightarrow \dfrac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\dfrac{1}{2\left( x+1 \right)}+\dfrac{3x}{2\left( {{x}^{2}}+1 \right)}-\dfrac{1}{2\left( {{x}^{2}}+1 \right)} \\
\end{align}\]
We use the above equation and use it in right hand side of equation (1) to have;\[\begin{align}
  & dy=\left( \dfrac{1}{2\left( x+1 \right)}+\dfrac{3x}{2\left( {{x}^{2}}+1 \right)}-\dfrac{1}{2\left( {{x}^{2}}+1 \right)} \right)dx \\
 & \Rightarrow dy=\dfrac{1}{2\left( x+1 \right)}dx+\dfrac{3x}{2\left( {{x}^{2}}+1 \right)}dx-\dfrac{1}{2\left( {{x}^{2}}+1 \right)}dx \\
\end{align}\]
Let us integrate both sides using sum rule of integration with their respective variables to have;
\[\begin{align}
  & \int{dy}=\int{\dfrac{1}{2\left( x+1 \right)}dx}+\int{\dfrac{3x}{2\left( {{x}^{2}}+1 \right)}dx}-\int{\dfrac{1}{2\left( {{x}^{2}}+1 \right)}dx} \\
 & \Rightarrow y=\dfrac{1}{2}\int{\dfrac{1}{x+1}dx}+\dfrac{3}{2}\int{\dfrac{x}{{{x}^{2}}+1}dx}-\dfrac{1}{2}\int{\dfrac{1}{{{x}^{2}}+1}dx} \\
\end{align}\]
We integrate the first term at the right hand side using the standard integral $\int{\dfrac{dx}{x+a}}=\log \left( x+a \right)$ and the third term with the starred integral$\int{\dfrac{dx}{{{x}^{2}}+{{a}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)$. W have;
\[\begin{align}
  & \Rightarrow y=\dfrac{1}{2}\log \left( x+1 \right)+\dfrac{3}{2}\int{\dfrac{x}{{{x}^{2}}+1}dx}-\dfrac{1}{2}\times \dfrac{1}{1}{{\tan }^{-1}}\left( \dfrac{x}{1} \right) \\
 & \Rightarrow y=\dfrac{1}{2}\log \left( x+1 \right)+\dfrac{3}{2}\int{\dfrac{x}{{{x}^{2}}+1}dx}-\dfrac{1}{2}{{\tan }^{-1}}x+{{c}_{1}}.......\left( 4 \right) \\
\end{align}\]
We integrate the middle term using integration by substitution. Let us have $t={{x}^{2}}+1$ then we have by differentiation$dt=2xdx\Rightarrow dx=\dfrac{dt}{2x}$. So we have;
\[\dfrac{3}{2}\int{\dfrac{x}{{{x}^{2}}+1}}dx=\dfrac{3}{2}\int{\dfrac{x}{t}\times }\dfrac{dt}{2x}=\dfrac{3}{4}\int{\dfrac{dt}{t}=\dfrac{3}{4}}\log t=\dfrac{3}{4}\log \left( {{x}^{2}}+1 \right)\]
We put the above integral value sin equation (4) to have the general solution
\[y=\dfrac{1}{2}\log \left( x+1 \right)+\dfrac{3}{2}\log \left( {{x}^{2}}+1 \right)-\dfrac{1}{2}{{\tan }^{-1}}x+c\]
Here $c$ is a real constant of integration. We put the condition $y=1$when$x=0$ in the general solution to have ;
\[\begin{align}
  & 1=\dfrac{1}{2}\log \left( 0+1 \right)+\dfrac{3}{2}\log \left( {{0}^{2}}+1 \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( 0 \right)+c \\
 & \Rightarrow 1=0+0-0+c \\
 & \Rightarrow c=1 \\
\end{align}\]
We put $c=1$ in the general solution to have the particular solution as
\[y=\dfrac{1}{2}\log \left( x+1 \right)+\dfrac{3}{2}\log \left( {{x}^{2}}+1 \right)-\dfrac{1}{2}{{\tan }^{-1}}x+1\]

Note: We should note that a general solution always contains an arbitrary constant of integration $c$ but a particular solution does not. The particular solution is always obtained by putting a value of random contains in the general solution, if not it is called a singular solution. The degree of differential equation is 3 and the degree of solution equation is 2 which verify our result since the given differential equation is of first order.