- #1

karush

Gold Member

MHB

- 3,269

- 5

$\tiny{2.1.{7}}$

$$\displaystyle y^\prime +y =\frac{1}{1+x^2}, \quad y(0)=0$$

$\textit{Find the solution of the given initial value problem.}$ \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

$$\displaystyle y^\prime +y =\frac{1}{1+x^2}, \quad y(0)=0$$

$\textit{Find the solution of the given initial value problem.}$ \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

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