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.

34. ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$, where $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\{\begin{array}{c}0,x<0\\ cosx,0⩽x⩽4\pi \\ 0,x>4\pi \end{array}\begin{array}{}\end{array}$

Short answer

$y=\sin x+\left\{\begin{array}{cc}0,& x<0\\ \frac{1}{2}x\sin x,& 0⩽x⩽4\pi \\ 2\pi \sin x,& x>4\pi \end{array}\right.$

Step-by-step solution

To find the solutions of the corresponding homogeneous differential equation

We have the second order differential equation

$y^{\text{'}\text{'}}+y=\left\{\begin{array}{cc}0,& x<0\\ \cos x,& 0⩽x⩽4\pi \\ 0,& x>4\pi \end{array}\right.$

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

$y\left(0\right)=0\text{and}{y}^{\text{'}}\left(0\right)=1$

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

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

$y^{\text{'}\text{'}}+y=0$ ……. (1)

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

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

After that, substitute with $y=e^{mx}$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

$m^2+1=0$

Then the roots are

$m_{1,2}=\pm i$

Which are conjugate complex.

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

$y_c=k_1\cos x+k_2\sin x$ ….. (a)

We have to apply the initial condition of the given differential equation as the following:

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

$$\begin{gathered} 0=c_1\cos \left(0\right)+c_2\sin \left(0\right) \\ {c}_1=0 \end{gathered}$$ …… (g)

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

Then apply the point into this first derivative function as

$$\begin{gathered} 1=-c_1\sin 0+c_2\cos 0 \\ {c}_2=1 \end{gathered}$$ …… (h)

Then, substitute with and shown in equations and into equation, then can we obtain the solution of the homogeneous D.E as

$y_h=\sin x$ ……. (3)

To find the particular solution yp(x) of the differential equation as the following technique:

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 $y_1=\cos xand{y}_2=\sin x$, then we can obtain the wronskian $W\left(y_1,y_2\right)$as

$$\begin{gathered} W(\cos x,\sin x)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right| \\ =\left|\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right| \\ =\left(\cos x\right)\times \left(\cos x\right)-\left(\sin x\right)\times (-\sin x) \\ ={\cos }^{2}x+{\sin }^{2}x \\ =1 \end{gathered}$$

$$\begin{gathered} W(\cos x,\sin x)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right| \\ =\left|\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right| \\ =\left(\cos x\right)\times \left(\cos x\right)-\left(\sin x\right)\times (-\sin x) \\ ={\cos }^{2}x+{\sin }^{2}x \\ =1 \end{gathered}$$

To find the Green’s function for the given equation

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

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

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

For the first interval,, we have, then we can obtain its particular solution as

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

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

For the second interval, $0⩽x⩽4\pi$, we have, $f\left(t\right)=\cos t$then we can obtain its particular solution $y_{p2}\left(x\right)$as

$$\begin{gathered} y_{p2}\left(x\right)={\int }_{x_0}^{x}G(x,t)f\left(t\right)dt \\ ={\int }_0^x(\sin x\cos t-\cos x\sin t)\left(\cos t\right)dt \\ =\sin x{\int }_0^x{\cos }^{2}tdt-\cos x{\int }_0^x\sin t\cos tdt \\ {\left.=\sin x{\int }_0^x\left(\frac{1}{2}+\frac{1}{2}\cos 2t\right)dt-\cos x\right]}_0^x\sin t\cos tdt \\ =\sin x{\left[\frac{1}{2}t+\frac{1}{4}\sin 2t\right]}_0^x-\cos x\left[\frac{1}{2}{\sin }^{2}t\right]{ \\ }_0^x=\sin x{\left[\frac{1}{2}t+\frac{1}{4}\left(2\sin t\cos t\right)\right]}_0^x-\cos x{\left[\frac{1}{2}{\sin }^{2}t\right]}_0^x \\ =\sin x{\left[\left(\frac{1}{2}x+\frac{1}{2}\sin x\cos x\right)-\left(0\right)\right]}^2\cos x\left[\left(\frac{1}{2}{\sin }^{2}x\right)-\left(0\right)\right] \\ =\frac{1}{2}x\sin x+\frac{1}{2}{\sin }^{2}x\cos x-\frac{1}{2}{\sin }^{2}x\cos x \\ =\frac{1}{2}x\sin x \end{gathered}$$

For the third interval, $x<3\pi$, we have, $f\left(t\right)=0$then we can obtain its particular solution $y_{p1}\left(x\right)$as

$$\begin{gathered} y_{p3}\left(x\right)={\int }_{x_0}^{\pi }G(x,t)f\left(t\right)dt \\ ={\int }_0^{4\pi }(\sin x\cos t-\cos x\sin t)\left(\cos t\right)dt-{\int }_{4\pi }^x(\sin x\cos t-\cos x\sin t)\left(0\right)dt \\ ={\int }_0^{4\pi }(\sin x\cos t-\cos x\sin t)\left(10\right)dt-0 \\ =\sin x{\left[\frac{1}{2}t+\frac{1}{4}\left(2\sin t\cos t\right)\right]}_0^{4\pi }-\cos x{\left[\frac{1}{2}{\sin }^{2}t\right]}_0^{4\pi } \\ =\sin x\left[\left(\frac{1}{2}4\pi +\frac{1}{2}\sin 4\pi \cos 4\pi \right)-\left(0\right)\right]-\cos x\left[\left(\frac{1}{2}{\sin }^{2}4\pi \right)-\left(0\right)\right] \\ =\sin x\left[\right(2\pi +0)-(0\left)\right]-\cos x\left[\right(0)-(0\left)\right] \\ =2\pi \sin x \end{gathered}$$

Final proof

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

$y_p\left(x\right)=\left\{\begin{array}{cc}0& x<0\\ \frac{1}{2}x\sin x,& 0⩽x⩽4\pi \\ 2\pi \sin x,& x>4\pi \end{array}\right.$ ……. (4)

Then from (3) and (4), we can obtain the solution of the given differential equation as

$$\begin{gathered} y=y_h+y_p \\ =\sin x+\left\{\begin{array}{cc}0& x<0\\ \frac{1}{2}x\sin x,& 0⩽x⩽4\pi \\ 2\pi \sin x,& x>4\pi \end{array}\right. \end{gathered}$$

The result is

$y=\sin x+\left\{\begin{array}{cc}0& x<0\\ \frac{1}{2}x\sin x,& 0⩽x⩽4\pi \\ 2\pi \sin x,& x>4\pi \end{array}\right.$