Question

In Problem 36 find a solution of the BVP when $\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{x}$.

Short answer

$y_p\left(x\right)=\frac{1}{6}\left(x^3-2x\right)$

Step-by-step solution

Given

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

To find the solutions of the corresponding homogeneous differential equation

We have the second order differential equation

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

with the initial boundary conditions

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

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 (2) 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

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

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 take the derivative for the homogeneous solution as

$$\begin{gathered} y^{\text{'}}=0+k_2 \\ =k_2 \end{gathered}$$

Then apply the condition$y\left(1\right)+y^{\text{'}}\left(1\right)=0$ into equation, then we have

$$\begin{gathered} 0=\left[k_1+k_2\left(1\right)\right]+\left[k_2\right] \\ {k}_1+2k_2=0 \\ {k}_1=-2k_2 \end{gathered}$$

If we choose $k_2=1$, then we have $k_1=-2$, then by substituting with $k_{1}and{k}_2$into equation we have

$$\begin{gathered} y_2=-2+x \\ =x-2 \end{gathered}$$

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 $y_1=xand{y}_2=x-2$and , then we can obtain the wronskian $W\left(y_1,y_2\right)$as

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

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

$$\begin{gathered} G(x,t)=\left\{\begin{array}{cc}\frac{y_1\left(t\right)y_2\left(x\right)}{W\left(t\right)},& 1⩽t⩽x\\ \frac{y_1\left(x\right)y_2\left(t\right)}{W\left(t\right)},& x⩽t⩽3\end{array}\right. \\ =\left\{\begin{array}{cc}\frac{\left(t\right)(x-2)}{2},& 0⩽t⩽x\\ \frac{\left(x\right)(t-2)}{2},& x⩽t⩽1\end{array}\right. \\ =\left\{\begin{array}{cc}\frac{1}{2}t(x-2),& 0⩽t⩽x\\ \frac{1}{2}x(t-2),& x⩽t⩽1\end{array}\right. \end{gathered}$$

Final proof

Now since we have, then we can obtain the particular solution for the given differential equation as

$$\begin{gathered} y_p\left(x\right)={\int }_0^{1}G(x,t)f\left(t\right)dt \\ ={\int }_0^{1}G(x,t)tdt \\ ={\int }_0^x\frac{1}{2}t(x-2)tdt+{\int }_x^1\frac{1}{2}x(t-2)tdt \\ =\frac{1}{2}(x-2){\int }_0^x{t}^{2}dt+\frac{1}{2}x{\int }_x^1\left(t^2-2t\right)dt \\ =\frac{1}{2}(x-2){\left[\frac{1}{3}{t}^3\right]}_0^x+\frac{1}{2}x{\left[\frac{1}{3}{t}^3-t^2\right]}_x^1 \end{gathered}$$

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

$$\begin{gathered} =\frac{1}{6}{x}^4-\frac{1}{3}{x}^3-\frac{1}{3}x-\frac{1}{6}{x}^4+\frac{1}{2}{x}^3 \\ =\frac{1}{6}{x}^3-\frac{1}{3}x \\ =\frac{1}{6}\left(x^3-2x\right) \end{gathered}$$

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

The result is

$y_p\left(x\right)=\frac{1}{6}\left(x^3-2x\right)$$y_p\left(x\right)=\frac{1}{6}\left(x^3-2x\right)$