Question

In Problems 35 and 36, (a) use (27) and (28) to find a solution of the boundary-value problem. (b) Verify that the function ${\mathit{y}}_{\mathbf{p}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ satisfies the differential equations and both boundary-conditions.

$\mathbf{36}\mathbf{.}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

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

Step-by-step solution

To find the solutions of the corresponding homogeneous differential equation

We have the second order differential equation

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

With the initial boundary conditions

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

and we have to get its boundary value problem as the following technique :

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

$y^{\text{'}\text{'}}=0$ …… (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 equations and (3) into the differential equation (1), then we obtain

$m^2{e}^{mx}=0$

Since$e^{mx}$cannot be equal 0, the we have

$m^2=0$

Then the roots are

$m_{1,2}=0$

Which are real and repeated.

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

$y_c=k_1+k_{2}x$ ….. (a)

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

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

$0=k_1+k_2\times \left(0\right)k_1=0$

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

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

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

$$\begin{gathered} 0=k_1+k_2\left(1\right) \\ {k}_1+k_2=0 \end{gathered}$$

If we choose $k_2=1$, then we have $k_1=-1$, then by substituting with $k_1$and$k_2$ into equation we have

$y_2=x-1$

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

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

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

$$\begin{gathered} G(x,t)=\left\{\begin{array}{cc}\frac{y_1\left(t\right)y_2\left(x\right)}{W\left(t\right)},& 1⩽t⩽x\\ \frac{y_1\left(x\right)y_2\left(t\right)}{W\left(t\right)},& x⩽t⩽3\end{array}\right. \\ =\left\{\begin{array}{cc}\frac{\left(t\right)(x-1)}{1},& 0⩽t⩽x\\ \frac{\left(x\right)(t-1)}{1},& x⩽t⩽1\end{array}\right. \\ =\left\{\begin{array}{cc}t(x-1),& 0⩽t⩽x\\ x(t-1),& x⩽t⩽1\end{array}\right. \end{gathered}$$

Final proof

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

$$\begin{gathered} y_p\left(x\right)={\int }_0^{1}G(x,t)f\left(t\right)dt \\ ={\int }_0^{1}G(x,t)\left(1\right)dt \\ ={\int }_0^{x}t(x-1)dt+{\int }_x^{1}x(t-1)dt \\ =(x-1){\int }_0^{x}tdt+x{\int }_x^1(t-1)dt \\ =(x-1){\left[\frac{1}{2}{t}^2\right]}_0^x+x{\left[\frac{1}{2}{t}^2-t\right]}_x^1 \end{gathered}$$

$$\begin{gathered} =(x-1)\left[\frac{1}{2}{x}^2-0\right]+x\left[\left(\frac{1}{2}-1\right)-\left(\frac{1}{2}{x}^2-x\right)\right] \\ =(x-1)\left[\frac{1}{2}{x}^2-0\right]+x\left[-\frac{1}{2}-\frac{1}{2}{x}^2+x\right] \\ =\frac{1}{2}{x}^3-\frac{1}{2}{x}^2-\frac{1}{2}x-\frac{1}{2}{x}^3+x^2 \\ =\frac{1}{2}{x}^2-\frac{1}{2}x \end{gathered}$$

The result is

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