2.1.7 Find the solution of the given initial value problem.

[karush]

$\tiny{2.1.{7}}$
$$\displaystyle y^\prime +y =\frac{1}{1+x^2}, \quad y(0)=0$$
$\textit{Find the solution of the given initial value problem.}$ \begin{align*}\displaystyle
u(x) &=e^x\\
(e^x y)'&=\frac{e^x}{1+x^2} \\
e^x y&=\int \frac{e^x}{1+x^2}\, dx\\
%\textit{book answer}
&=\color{red}
{\displaystyle e^{-x}\int_{0}^{x}\frac{e^t}{1+t^2} \, dt}
\end{align*}
need help with steps why does the answer have $t$ in it
also if y=0 wouldn't it be 1

 

[MarkFL]

The integrand:

\(\displaystyle \frac{e^x}{x^2+1}\)

does not have an anti-derivative in elementary terms, and so the solution is given as a definite integral. The dummy variable $t$ is introduced so that a different variable is used than what is in the upper limit in integration, for the sake of good notation.