Question

In Problems 31 and 32 solve the given initial-value problem.

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$, where$$\begin{gathered} \mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\left\{e^{-x},0<x<1 \\ 0,x⩾1\right\} \end{gathered}$$

Short answer

The solution is$$\begin{gathered} y\left(x\right)=\left\{5e^{-x}+e^{-x}x,0⩽x<1 \\ 6e^{-x},x⩾1\right\} \end{gathered}$$

Step-by-step solution

Definition

The standard form of a linear differential equation is $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\mathbf{+}\mathit{P}\mathit{y}\mathbf{+}\mathit{Q}$, and it contains the variable$\mathbf{y}$, and its derivatives.

Find Solution

On comparing we have$P\left(x\right)=1\text{and}f\left(x\right)=e^{-x}$

Now, substitute the value$P\left(x\right)$ and$f\left(x\right)$ in the Solution formula

$y\left(x\right)=e^{-\int P\left(x\right)\text{d}x}\left[c+\int f\left(x\right)e^{\int P\left(x\right)\text{d}x}\text{d}x\right]$

$$\begin{gathered} y\left(x\right)=e^{-\int \text{d}x}\left[c+\int e^{-x}{e}^{\int \text{d}x}\text{d}x\right] \\ y\left(x\right)=e^{-x}\left[c+\int e^{-x}{e}^x\text{d}x\right] \\ y\left(x\right)=e^{-x}[c+x] \\ y\left(x\right)=e^{-x}c+x{e}^{-x} ------\left(1\right) \end{gathered}$$

Apply initial condition

Use the initial condition $y\left(0\right)=5$, to get $c$.

$\begin{array}{rcl}y\left(0\right)& =& e^{0}c+0\left(e^0=1\right)\\ 5& =& c\end{array}$

Substitute the value$c=5$ in the equation (1).

$y\left(x\right)=5e^{-x}+x{e}^{-x}$----(2)

Integrate for x⩾1

Now is considering $x⩾1$, for which $f\left(x\right)=0$. So, the given differential equation, can be written asrole="math" localid="1663910091943" $\frac{dy}{dx}+y=0$

Linear differential equation of the first order $y^{\text{'}}+P\left(x\right)y=f\left(x\right)$

So, we have$P\left(x\right)=1$ and$f\left(x\right)=0$

Now, substitute the value$P\left(x\right)$ and $f\left(x\right)$in the formula$y\left(x\right)=e^{-\int P\left(x\right)\text{d}x}\left[c+\int f\left(x\right)e^{\int P\left(x\right)\text{d}x}\text{d}x\right]$ to get$y\left(x\right)=e^{-\int \text{d}x}\left[c+\int 0·e^{\int \text{d}x}\text{d}x\right]$

$y\left(x\right)=e^{-x}c$----(3)

Apply initial condition

Substitute the value$x=1$ in the equation (2).

$$\begin{gathered} y\left(1\right)=5e^{-1}+e^{-1} \\ y\left(1\right)=6e^{-1} \end{gathered}$$

And substitute the value$x=1$ in the equation (3).

$y\left(1\right)=c{e}^{-1}$

So, we get$c=6$

Substitute the value$c=6$ in the equation (3) we have:$y\left(x\right)=6e^{-x}$

Therefore the solution of given differential equation is $$\begin{gathered} y\left(x\right)=\left\{5e^{-x}+e^{-x}x,0⩽x<1 \\ 6e^{-x},x⩾1\right\} \end{gathered}$$.