The problem says:

Solve by variation of parameters method: $\ y''-12y'+y=e^{6x}\ln(x) $

I am having trouble obtaining the particular solution. Is the Wronskian necessary for this problem? I'm just not sure what the particular solution should look like.

  • $\begingroup$ You find the $\ y_H=c_1y_2+c_2y_2$ then you say that $\ c_1=c_1(x),c_2=c_2(x)$ $\ y_H$ is the result of the equivalent homogenus equation then you solve the system $\ c_1'(x)y_1+c_2'(x)y_2=0\ and\ c_1'(x)y'_1+c_2'(x)y'_2=e^{6x}ln(x)$ $\endgroup$
    – CTSnake
    May 2 '17 at 16:47
  • $\begingroup$ The solution will contain non-elementary functions. $\endgroup$ May 2 '17 at 17:28

I assume you know how to find the complementary solution. It should evaluate as: $$y_c(x)=c_1 e^{(6-\sqrt{35})x}+c_2 e^{(6+\sqrt{35})x}$$

This is a second-order linear nonhomogeneous differential equation given by: $$y''+p(x)y'+q(x)=r(x)$$

You should indeed be using the Wronskian. The basis solutions are $e^{(6-\sqrt{35})x}$ and $e^{(6+\sqrt{35})x}$. Therefore: $$W(x)=\begin{vmatrix} e^{(6-\sqrt{35})x} & e^{(6+\sqrt{35})x} \\ (6-\sqrt{35})e^{(6-\sqrt{35})x} & (6+\sqrt{35})e^{(6+\sqrt{35})x} \end{vmatrix}=2\sqrt{35}e^{12x}$$ The particular solution should be given by: $$y_p(x)=v_1(x)\cdot e^{(6-\sqrt{35})x}+v_2(x)\cdot e^{(6+\sqrt{35})x} \tag{1}$$ Where: $$v_1(x)=-\int \frac{r(x)\cdot e^{(6+\sqrt{35})x}}{W(x)}~dx=-\int \frac{e^{\sqrt{35}\cdot x}\cdot \ln(x)}{2\sqrt{35}}~dx$$ And: $$v_2(x)=\int \frac{r(x)\cdot e^{(6-\sqrt{35})x}}{W(x)}~dx=\int \frac{e^{-\sqrt{35}\cdot x}\cdot \ln(x)}{2\sqrt{35}}~dx$$ Integrating $v_1(x)$ and $v_2(x)$ is not possible using a finite number of elementary functions, one must express it in terms of the Exponential Integral.

The exponential integral $\operatorname*{Ei}(x)$ is defined as: $$\operatorname{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}{t}~dt \tag{2}$$

One can use this definition to show that: $$\int \frac{e^x}{x}~dx=\operatorname*{Ei}(x)+C \tag{3}$$ We will be using this result in the next section.

I will compute $v_1(x)$ for you: Substitute $u=\sqrt{35}\cdot x$. Integrate by parts, and use $(3)$:

$$\begin{align}-\int \frac{e^{-\sqrt{35}\cdot x}\cdot \ln(x)}{2\sqrt{35}}~dx&=-\frac{1}{70}\int e^u\cdot \ln\left(\frac{u}{\sqrt{35}}\right)~du\\&=-\frac{1}{70}\left(e^u \ln\left(\frac{u}{\sqrt{35}}\right)-\int \frac{e^u}{u}~du\right)\\&=-\frac{1}{70}\left(e^u \ln\left(\frac{u}{\sqrt{35}}\right)-\operatorname*{Ei}(u)\right)+C\\&=\frac{1}{70}\left(\operatorname*{Ei}(\sqrt{35}\cdot x)-e^{\sqrt{35}\cdot x}\cdot \ln(x)\right)+C\end{align}$$ Therefore, we have: $$v_1(x)=\frac{1}{70}\left(\operatorname*{Ei}(\sqrt{35}\cdot x)-e^{\sqrt{35}\cdot x}\cdot \ln(x)\right)$$ Note that the constant of integration was purposefully omitted (I suppose you know why).

One can similarly evaluate $v_2(x)$ using the substitution $v=-\sqrt{35}\cdot x$.

Hopefully, you can evaluate $v_2(x)$ and substitute it all into $(1)$ to obtain the particular solution. Then, you can use the following to give you the general solution: $$y(x)=y_c(x)+y_p(x)$$

  • $\begingroup$ The constant of integration was omitted because if you substitute it into your particular solution, you obtain: $$\begin{align}y_p(x)&=\left(\frac{1}{70}\left(\operatorname*{Ei}(\sqrt{35}\cdot x)-e^{\sqrt{35}\cdot x}\cdot \ln(x)\right)+C\right)e^{(6-\sqrt{35})x}+\cdots\\&=\frac{e^{(6-\sqrt{35})x}}{70}\left(\operatorname*{Ei}(\sqrt{35}\cdot x)-e^{\sqrt{35}\cdot x}\cdot \ln(x)\right)+\color{red}{Ce^{(6-\sqrt{35})x}}+\cdots \end{align}$$ Then, if you use the fact that $y(x)=y_c(x)+y_p(x)$: we can let $k=c_1+C$, we get a new arbitrary constant and thus this is why we omit it as shown below: $\endgroup$ May 2 '17 at 18:18
  • $\begingroup$ $$\begin{align}y(x)&=y_c(x)+y_p(x)\\&=\color{red}{c_1 e^{(6-\sqrt{35})x}}+c_2 e^{(6-\sqrt{35})x}+\frac{e^{(6-\sqrt{35})x}}{70}\left(\operatorname*{Ei}(\sqrt{35}\cdot x)-e^{\sqrt{35}\cdot x}\cdot \ln(x)\right)+\color{red}{Ce^{(6-\sqrt{35})x}}+\cdots\\&=\color{red}{k e^{(6-\sqrt{35})x}}+c_2 e^{(6-\sqrt{35})x}+\frac{e^{(6-\sqrt{35})x}}{70}\left(\operatorname*{Ei}(\sqrt{35}\cdot x)-e^{\sqrt{35}\cdot x}\cdot \ln(x)\right)+\cdots \end{align}$$ The same applies for $v_2(x)$, we can omit the constant of integration. $\endgroup$ May 2 '17 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.