Question

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

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{P}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{\lambda }}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$,whererole="math" localid="1663910909848" $$\begin{gathered} \mathit{P}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\left\{1,0⩽x<1 \\ -1,x⩾1\right\} \end{gathered}$$

Short answer

The solution is$$\begin{gathered} y\left(x\right)=\left\{-\frac{3}{2}{e}^{-x}+\frac{1}{2}{e}^x,0⩽x<1 \\ -\frac{3}{2}{e}^{-2}{e}^x-\frac{1}{2}{e}^x+x{e}^x,x⩾1\right\} \end{gathered}$$

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 $\mathit{y}$, and its derivatives.

Find Solution

On comparing with general linear differential equation 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 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

$$\begin{gathered} y\left(x\right)=e^{-\int \text{d}x}\left[c+\int e^x{e}^{\int \text{d}x}\text{d}x\right] \\ y\left(x\right)=e^{-x}\left[c+\int e^x{e}^x\text{d}x\right] \\ y\left(x\right)=e^{-x}\left[c+\frac{1}{2}{e}^{2x}\right] \\ y\left(x\right)=e^{-x}c+\frac{1}{2}{e}^x ----\left(1\right) \end{gathered}$$

Apply initial condition

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

role="math" localid="1663911233319" $\begin{array}{rcl}y\left(0\right)& =& e^{0}c+\frac{1}{2}{e}^0\left(e^0=1\right)\\ -1& =& c+\frac{1}{2}\\ c& =& -\frac{3}{2}\end{array}$

Substitute the value $c=-\frac{3}{2}$in the equation (1).

$y\left(x\right)=-\frac{3}{2}{e}^{-x}+\frac{1}{2}{e}^x ---\left(2\right)$

Integrate for x⩾1

For$x⩾1$, given$P\left(x\right)=-1$ . So, the given differential equation, can be written as

$\frac{dy}{dx}-y=e^x$

So, 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 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

$$\begin{gathered} y\left(x\right)=e^{\int \text{d}x}\left[c+\int e^x{e}^{-\int \text{d}x}\text{d}x\right] \\ y\left(x\right)=e^x\left[c+\int e^x{e}^{-x}\text{d}x\right] \\ y\left(x\right)=e^x[c+x] \\ y\left(x\right)=e^{x}c+x{e}^x ----\left(3\right) \end{gathered}$$

Apply initial condition

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

$y\left(1\right)=-\frac{3}{2}{e}^{-1}+\frac{1}{2}e$

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

$y\left(1\right)=ec+e$

So, we have

$\begin{array}{rcl}-\frac{3}{2}{e}^{-1}+\frac{1}{2}e& =& ec+e\\ -\frac{3}{2}{e}^{-1}+\frac{1}{2}e-e& =& ec\\ -\frac{3}{2}{e}^{-1}-\frac{1}{2}e& =& ec\\ -\frac{3}{2}{e}^{-2}-\frac{1}{2}& =& c\end{array}$

Substitute the value $c=-\frac{3}{2}{e}^{-2}-\frac{1}{2}$in the equation (3) we have

$y\left(x\right)=-\frac{3}{2}{e}^{-2}{e}^x-\frac{1}{2}{e}^x+x{e}^x$