Question

In Problems 39-44 proceed as in Examples 7 and 8 to find a solution of the given boundary-value problem.

39. ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$y_p=1-\cos x+\left(\frac{\cos 1-{\sin }^{2}1-\sin 1\cos 1}{\sin 1}\right)\sin x$

Step-by-step solution

To find the solutions of the corresponding homogeneous differential equation

We have the second order differential equation

$y^{\text{'}\text{'}}+y=1$

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 boundary conditions

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

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 x, then we have

$$\begin{gathered} y^{\text{'}}=m{e}^{mx} \\ {y}^{\text{'}\text{'}}=m^2{e}^{mx} \end{gathered}$$ …. (2) & (3)

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)

Now we have to apply the initial boundary condition of the given differential equation as the following:

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

$$\begin{gathered} 0=k_1\cos \left(0\right)+k_2\sin \left(0\right) \\ {k}_1=0 \end{gathered}$$

Let $k_2=1$, then substitute with$k_1=0$ and$k_2=1$ into equation, then we have

$$\begin{gathered} y_1=0+\sin x \\ =\sin x \end{gathered}$$

After that, we have to apply the point$(x,y)=(1,0)$ into equation, then we have

$0=k_1\cos \left(1\right)+k_2\sin \left(1\right)$

Let $k_1=1$, then we have

$$\begin{gathered} 0=\left(1\right)\cos \left(1\right)+k_2\sin \left(1\right) \\ {k}_2\sin \left(1\right)=-\cos \left(1\right) \\ {k}_2=-\cot \left(1\right) \end{gathered}$$

Then, substitute with$k_1=1and{k}_2=-\cot \left(1\right)$ into equation , then we have

$y_2=\cos x-\cot \left(1\right)\sin x$

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 $y_1=\sin xand{y}_2=\cos x-\cot 1\sin x$, then we can obtain the wronskian $W\left(y_1,y_2\right)$as

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

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

$G(x,t)=\left\{\begin{array}{cc}\frac{y_1\left(t\right)y_2\left(x\right)}{W\left(t\right)},& 0⩽t⩽x\\ \frac{y_1\left(x\right)y_2\left(t\right)}{W\left(t\right)},& x⩽t⩽1\end{array}\right.$

$$\begin{gathered} =\left\{\begin{array}{cc}\frac{\sin t(\cos x-\cot 1\sin x)}{-1},& 0⩽t⩽x\\ \frac{\sin x(\cos t-\cot 1\sin t)}{-1},& x⩽t⩽1\end{array}\right. \\ =\left\{\begin{array}{cc}\sin t(-\cos x+\cot 1\sin x),& 0⩽t⩽x\\ \sin x(-\cos t+\cot 1\sin t),& x⩽t⩽1\end{array}\right. \end{gathered}$$

Final proof

Now since we have $f\left(t\right)=1$, then we can obtain the particular solution$y_p\left(x\right)$ as

$$\begin{gathered} y_p\left(x\right)={\int }_0^{1}G(x,t)f\left(t\right)dt \\ ={\int }_0^{1}G(x,t)\left(1\right)dt \\ ={\int }_0^x\sin t(-\cos x+\cot 1\sin x)dt+{\int }_x^1\sin x(-\cos t+\cot 1\sin t)dt \\ =(-\cos x+\cot 1\sin x){\int }_0^x\sin tdt+\sin x{\int }_x^1(-\cos t+\cot 1\sin t)dt \\ =(-\cos x+\cot 1\sin x)[-\cos t{]}_0^x+\sin x[-\sin t-\cot 1\cos t{]}_x^1 \\ =(-\cos x+\cot 1\sin x)[-\cos x+\cos 0]+\sin x\left[\right(-\sin 1-\cot 1\cos 1)-(-\sin x-\cot 1\cos x\left)\right] \end{gathered}$$

$$\begin{gathered} y_p=\left({\cos }^{2}x-\cos x-\cot 1\sin x\cos x+\cot 1\sin x\right] \\ +\left(-\sin 1\sin x-\cot 1\cos 1\sin x+{\sin }^{2}x+\cot 1\sin x\cos x\right) \\ =1-\cos x+\cot 1\sin x-\sin 1\sin x-\cot 1\cos 1\sin x \\ =1-\cos x+(\cot 1-\sin 1-\cos 1)\sin x \\ =1-\cos x+\left(\frac{\cos 1-{\sin }^{2}1-\sin 1\cos 1}{\sin 1}\right)\sin x \end{gathered}$$

The final result is

$y_p=1-\cos x+\left(\frac{\cos 1-{\sin }^{2}1-\sin 1\cos 1}{\sin 1}\right)\sin x$