Question

Proceed as in Example 7 and express the solution of the given initial value problem in terms of an integral defined function.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

So, the solution of the given initial value problem in terms of an integral defined function$y=e^{-e^x{\int }_0^x{e}^{e^t}\mathrm{d}t+e^{1-e^x}}$

Step-by-step solution

Form of linear equation

The linear differential equation is of the form $\frac{\mathbf{dy}}{\mathbf{dx}}+\mathbf{Py}=\mathbf{Q}$,

where P and Qare numeric constants or functions in x.

Evaluation

Linear differential equation of the first order

$y^{\text{'}}+P\left(x\right)y=f\left(x\right)$

The integrating factor is given by

$e^{\int P\left(x\right)dx}$

Based on (1) and the given differential equation, we obtain

$P\left(x\right)=e^x\mathrm{and}f\left(x\right)=1$

Find the integrated factor

Use the formula (2), to get integrating factor

$\begin{array}{l}{e}^{\int P\left(x\right)dx}=e^{\int e^{x}dx}\\ =e^{e^x}\end{array}$

Thus, the solution is$y{e}^{e^x}={\int }_0^x{e}^{c^t}\mathrm{d}t+c$.

Use the initial condition $y\left(0\right)=1$to get c

role="math" localid="1664267914972" $\begin{array}{l}{e}^{e^0}=\underset{x\to 0}{lim}{\int }_0^x{e}^{e^0}dt+c\left(e^0-1\right)e\\ =e·0+cc\\ =e\end{array}$

Substitution

Substitute the value c =e in the equation (3)

$\begin{array}{l}y{e}^{e^x}={\int }_0^x{e}^{e^t}\mathrm{d}t+e\\ y=e^{-e^x{\int }_0^x{e}^{e^t}\mathrm{d}t+e^{1-e^x}}\end{array}$

So, the required solution is $y=e^{-e^x{\int }_0^x{e}^{e^t}\mathrm{d}t+e^{1-e^x}}$.