Soln of IVP: $y = \frac{e^x + 1 - e}{x}, \space x>0$

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This correction leads to the correct answer shown in (8).In summary, the solution to the initial value problem is given by $y=\frac{1}{x}(e^x+1-e)$ and the interval of validity is $x\neq 0$.
  • #1
karush
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$\textsf{ Find the solution of the given initial value problem.}$
$$xy^\prime+y=e^x, \qquad y(1)=1$$
$$\begin{array}{lrll}
\textit{Divide thru with x}\\
&\displaystyle y' +\frac{1}{x}y
&\displaystyle=\,\frac{e^x}{x} &_{(1)}\\
\textit {Find u(x)}\\
&\displaystyle u(x)
&\displaystyle=\exp\int\frac{1}{x}\,dx\\
&&=e^{\ln {x}}\\
&&=x &_{(2)}\\
\textit{Multiply thru with $x$} \\
&(xy)' +x'y&=e^x &_{(3)}\\
\textit{Rewrite:}\\
&(xy)'&=e^x &_{(4)}\\
\textit{Integrate }\\
&\displaystyle xy
&=\displaystyle\int e^x \, dx\\
&&=\displaystyle e^x+c &_{(5)}\\
\textit{Divide thru by $x$}\\
&\displaystyle y&=\displaystyle\frac{e^x}{x}+\frac{c}{e^x} &_{(6)}\\
\textit{So then if }\\
&\displaystyle y(1)&\displaystyle=e+\frac{c}{e}=1 &_{(7)}\\
\textit{with $c=?$ then }\\
&\displaystyle y
&=\color{red}{\displaystyle\frac{1}{x}(e^x + 1 - e)} &_{(8)}\\
\end{array}$$

ok (8) is the book answer but ? what wuld be c?
$\textit{State the interval in which the solution is valid. ?}\\$
 
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  • #2
Towards the end, when you divide through by $x$, you want:

\(\displaystyle y(x)=\frac{e^x}{x}+\frac{c}{x}\)

You mistakenly divided the constant by $e^x$.
 

FAQ: Soln of IVP: $y = \frac{e^x + 1 - e}{x}, \space x>0$

What is an initial value problem (IVP)?

An initial value problem is a type of differential equation that involves finding a function that satisfies both a given differential equation and a set of initial conditions. The initial conditions typically involve specifying the value of the function at a certain point or a certain value of the independent variable.

How do you solve an initial value problem?

To solve an initial value problem, you can use a variety of methods, such as separation of variables, substitution, or integrating factors. The specific method used depends on the form of the differential equation and the initial conditions given. In this case, the initial value problem can be solved using the method of separation of variables.

What is the solution to the given initial value problem?

The solution to the initial value problem $y = \frac{e^x + 1 - e}{x}, \space x>0$ is $y = e^x + C$, where $C$ is a constant determined by the initial condition given. In this case, the initial condition is $y(1) = 1$, so the solution is $y = e^x + 0 = e^x$.

Can you explain the significance of the restriction $x>0$?

The restriction $x>0$ means that the solution is only valid for positive values of $x$. This is because the given function is not defined for values of $x$ less than or equal to 0. Therefore, the solution to the initial value problem is only applicable for the positive domain of $x$.

What is the general solution to an initial value problem?

The general solution to an initial value problem is the most general form of the solution that satisfies the given differential equation and initial conditions. In this case, the general solution is $y = e^x + C$, where $C$ is a constant that can be determined by the initial condition given. It represents the family of all possible solutions to the initial value problem.

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