Question

In Problems 31–34 proceed as in Example 6 to ­find a solution of the initial-value problem with the given piecewise-de­fined forcing function.

32.$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$, where$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}0,x<0\\ x,x⩾0\end{array}$

Short answer

$y=5e^x+3e^{-x}+\left\{\begin{array}{l}1-\cosh x,x<0\\ \cosh x-1,x⩾0\end{array}\right.$

Step-by-step solution

The second order differential equation

$y=e^{mx}$We have the second order differential equation

$y^{\text{'}\text{'}}-y=\left\{\begin{array}{l}0,x<0\\ x,x⩾0\end{array}\right.$

as the form

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

with the initial conditions

$y\left(0\right)=3\text{and}{y}^{\text{'}}\left(0\right)=2$

and we have to get its solution as the following technique:

First, we have to find the solutions of the corresponding homogeneous differential equation

$y^{\text{'}\text{'}}-y=0$

by assuming that$y=e^{mx}$is a solution for this homogeneous differential equation, then follow then differentiate this assumption with respect to x, then we have

$$\begin{gathered} y^{\text{'}}=m{e}^{mx} \\ {y}^{\text{'}\text{'}}=m^2{e}^{mx} \end{gathered}$$

After that, substitute with and equation (2) into the differential equation (1), then we obtain

$$\begin{gathered} m^2{e}^{mx}-e^{mx}=0 \\ \left(m^2-1\right)e^{mx}=0 \end{gathered}$$

Since $x^m$cannot be equal 0, the we have

$$\begin{gathered} m^2-1=0 \\ (m-1)(m+1)=0 \end{gathered}$$

Then the roots are

$m_1=1\text{and}{m}_2=-1$

which are real and distinct.

Then we can obtain the solution of the corresponding homogeneous differential equation as

$y_c=k_1{e}^x+k_2{e}^{-x}.........\left(a\right)$

Second, now we have to find the particular solution $y_p\left(x\right)$of the given differential equation as the following technique:

Since we have $y_1=e^{x}and{y}_2=e^{-x}$, then we can obtain the wronskian$W\left(y_1,y_2\right)$ as

$$\begin{gathered} W\left(e^x,e^{-x}\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right| \\ =\left|\begin{array}{cc}{e}^x& e^{-x}\\ e^x& -e^x\end{array}\right| \\ =\left(e^x\right)\times \left(-e^{-x}\right)-\left(e^{-x}\right)\times \left(e^x\right) \\ =-e^0-e^0 \\ =-2 \end{gathered}$$

The given differential equation

After that, we have to find the Green's function for the given differential equation as

$$\begin{gathered} G(x,t)=\frac{y_1\left(t\right)y_2\left(x\right)-y_1\left(x\right)y_2\left(t\right)}{W\left(t\right)} \\ =\frac{e^t\times e^{-x}-e^{-t}\times e^x}{-2} \\ =\frac{e^x\times e^{-t}-e^{-x}\times e^t}{2} \\ =\frac{1}{2}{e}^x{e}^{-t}-\frac{1}{2}{e}^{-x}{e}^t \end{gathered}$$

Now since we have two different intervals for the particular solution, then we have to obtain it$y_p\left(x\right)$as the following:

For $x<0$, we have $f\left(t\right)=0$, then we can obtain its particular solution $y_{p1}\left(x\right)$as

$$\begin{gathered} y_{p1}\left(x\right)={\int }_{x_0}^{x}G(x,t)f\left(t\right)dt \\ ={\int }_0^x\left(\frac{1}{2}{e}^x{e}^{-t}-\frac{1}{2}{e}^{-x}{e}^t\right)\left(0\right)dt \\ =0 \end{gathered}$$

For $x⩾0$, we have $f\left(t\right)=t$, then we can obtain its particular solution$y_{p2}\left(x\right)$as

$$\begin{gathered} y_{p2}\left(x\right)={\int }_{x_0}^{x}G(x,t)f\left(t\right)dt \\ ={\int }_0^x\left(\frac{1}{2}{e}^x{e}^{-t}-\frac{1}{2}{e}^{-x}{e}^t\right)\left(t\right)dt \\ =\frac{1}{2}{e}^x{\int }_0^{x}t{e}^{-t}dt-\frac{1}{2}{e}^{-x}{\int }_0^{x}t{e}^{t}dt \end{gathered}$$

The integration$\int t{e}^{-t}dt$can be made using integration by parts as the following:

If we set $s=t$, then $ds=dt$, and if we set $dv=e^{-t}dt$, then $v=-e^{-t}$, then we obtain

$$\begin{gathered} \int t{e}^{-t}dt=s\times v-\int vds \\ =-t{e}^{-t}+\int e^{-t}dt \\ =-t{e}^{-t}-e^{-t} \end{gathered}$$

and the integration can be made using integration by parts as the following:

If we set $s=t$, then $ds=dt$, and if we set $dv=e^{t}dt$, then $v=e^t$, then we obtain

$$\begin{gathered} \int t{e}^{t}dt=s\times v-\int vds \\ =t{e}^t-\int e^{t}dt \\ =t{e}^t-e^t \end{gathered}$$

Then we can obtain$y_{p2}\left(x\right)$as

$$\begin{gathered} y_{p2}\left(x\right)=\frac{1}{2}{e}^x{\left[-t{e}^{-t}-e^{-t}\right]}_0^x-\frac{1}{2}{e}^{-x}{\left[t{e}^t-e^t\right]}_0^x \\ =\frac{1}{2}{e}^x\left[\left(-x{e}^{-x}-e^{-x}\right)-(0-1)\right]-\frac{1}{2}{e}^{-x}\left[\left(x{e}^x-e^x\right)-(0-1)\right] \\ =-\frac{1}{2}x-\frac{1}{2}+\frac{1}{2}{e}^x-\frac{1}{2}x+\frac{1}{2}-\frac{1}{2}{e}^{-x} \\ =\frac{e^x-e^{-x}}{2}-x \\ =\sinh x-x \end{gathered}$$

Then we can obtain the particular solution of the given differential equation as

$y_p\left(x\right)=\left\{\begin{array}{l}0,x<0\\ \sinh x-x,x⩾0\end{array}\right......\left(b\right)$

Then from (a) and (b), we can obtain the general solution of the given differential equation as

$$\begin{gathered} y=y_c+y_{p1}+y_{p2} \\ =k_1{e}^x+k_2{e}^{-x}+0+\sinh x-x \\ =k_1{e}^x+k_2{e}^{-x}+\sinh x-x...........\left(c\right) \end{gathered}$$

Finally, we have to apply the initial condition of the given differential equation as the following:

We have to apply the point$(x,y)=(0,3)$ into equation (c), then we have

$$\begin{gathered} 3=k_1{e}^0+k_2{e}^0+\sinh 0-0 \\ 3=k_1+k_2+0 \\ {k}_1+k_2=3..............\left(d\right) \end{gathered}$$

After that, we have to take the first derivative for the general solution as

$y^{\text{'}}=k_1{e}^x-k_2{e}^{-x}+\cosh x-1$

Then apply the point$\left(x,y^{\text{'}}\right)=(0,2)$ into this first derivative function as

$$\begin{gathered} 2=k_1{e}^0-k_2{e}^0+\cosh 0-1 \\ {k}_1-k_2=2..........\left(e\right) \end{gathered}$$

Final proof

Then by solving these two equations (d) and (e), we find that

$k_1=\frac{5}{2}\text{and}{k}_2=\frac{1}{2}$

Then, substitute with $k_{1}and{k}_2$into equation (c), then we obtain

$y=\frac{5}{2}{e}^x+\frac{1}{2}{e}^{-x}+\left\{\begin{array}{l}0,x<0\\ \sinh x-x,x⩾0\end{array}\right.$

is the solution of the given differential equation at the given conditions.