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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.

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  • $\begingroup$ 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
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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: $$y(x)=y_c(x)+y_p(x)$$

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  • $\begingroup$ 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

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