So, I was trying to solve this differential equation problem by using the "Wronskian method" even if I wasn't told to, but I'm used to do my problems with the previously mentioned method. Differential equation:


by using the method I get integrals, which I don't know how to solve:

$$∫\frac{1}{x^2e^{3x}}dx \quad \text{or} \quad ∫\frac{1}{xe^{3x}}dx$$

I even tried to figure out the solution by using some of those "online calculators", like wolfram alpha etc., but they use the so called "exponential integral E(U)", E1 or something like that, which we don't use on my math class. Is there any way to solve this differential equation? Feel free to post an explanation, covered with details. Peace

  • $\begingroup$ The particular solution should come out to $1/x$. $\endgroup$ – AlexanderJ93 Jan 18 '18 at 20:06
  • $\begingroup$ I get the following integrals, which are elementary: $$-\int \frac{9x^2+6x+2}{x^2 e^{3x}}~dx$$ $$\int \frac{9x^2+6x+2}{x^3 e^{3x}}~dx$$ Notice that those $\text{Ei}$ terms cancel out when you apply integration by parts. $\endgroup$ – projectilemotion Jan 18 '18 at 20:06
  • $\begingroup$ Hi! I get them as well, but what am i supposed to do now? From 1 integral to make 3 different ones or to use some kind of substitution for the whole thing? edit: well I don't know how to use those Ei terms. Is there any other way to avoid them? $\endgroup$ – akironjo Jan 18 '18 at 20:10
  • $\begingroup$ You don't need to know how to deal with them. Instead, notice that: $$\int \frac{9x^2+6x+2}{x^2e^{3x}}~dx=\int \frac{9}{e^{3x}}~dx+2\left(\int \frac{3}{xe^{3x}}~dx+\color{green}{\int\frac{1}{x^2e^{3x}}~dx}\right)$$ And note that via integration by parts: $$\color{green}{\int \frac{1}{x^2 e^{3x}}~dx}=-\frac{1}{xe^{3x}}-\int \frac{3}{xe^{3x}}~dx$$ Now, as I mentioned, those horrible integrals do cancel out. The other integral works in a similar way. $\endgroup$ – projectilemotion Jan 18 '18 at 20:21
  • $\begingroup$ well, you are basically telling me that I don't have to do the ∫3/(xe^(3x))dx, because it's going to "cancel" itself with same, negative one from the right side of the "green integral"? $\endgroup$ – akironjo Jan 18 '18 at 20:57

Hints (because the calculations are large, the exercise is easy) :

  • Find the complementary solution by solving : $$y'' - 6y' + 9y = 0$$ You can do this by assuming that a solution will be proportional to $y(x) =e^{\lambda x}$ and then proceed to finding an adequate $\lambda$ by plugging it in the equation above. Be careful about the multiplicity of the root yielded by the method above, as you'll need to take care of the order of your complementary solution. After the calculations (do them carefully), you should get :

$$y_c(x) = c_1e^{3x} + c_2e^{3x}x$$

  • Determine the particular soltuion of your initial differential equation $$y'' - 6y' + 9y = \frac{9x^2 + 6x + 2}{x^3}$$ by variation of parameters. You will need to compute the wronskian of the terms $y_1(x), y_2(x)$ of the solution $y(x)$ above and then find the integrals : $$v_1(x) = -\int \frac{f(x)y_2(x)}{W(x)}dx \quad \text{and} \quad v_2(x) = \int \frac{f(x)y_1(x)}{W(x)}dx$$ Then, the particular solution, will be : $$y_p(x) = v_1(x)y_1(x) + v_2(x)y_2(x)$$
  • Finally, the general solution is : $$y(x) = y_c(x) + y_p(x)$$

Note : This method applies to any such differential equation and is straight forward.

Side-note : You aren't going to get any "scary" integrals this way !

  • $\begingroup$ Hi! Thank you for your reply, but when I put everything in "v1(x)=−∫f(x)y2(x)W(x)dx" I get practically the same thing as before, or am I missing something out? v1(x)=−e^(3x)∫(9x^(2)+6x+2)/(e^(3x)*x^(2))dx ? $\endgroup$ – akironjo Jan 18 '18 at 20:47
  • $\begingroup$ @akironjo You should get : $$v_1(x) = - \int \frac{e^{-3x}(9x^2+6x+2)}{x^2}dx \quad \text{and} \quad v_2(x) = \int \frac{e^{-3x}(9x^2+6x+2)}{x^3}dx$$ Check your calculations as mentioned in my comment to proceed carefully. Note : The method elaborated works 100% smoothly as I've calculated the desired result (and it's also a really standard way of handling such differential problems). $\endgroup$ – Rebellos Jan 18 '18 at 21:06
  • $\begingroup$ Alright, but what I'm asking you is should I make three different integrals from the v1(x) by separating by numerators, because We have three different ones (if you know what I mean), or is there a substitution which I can use to cover the whole integral, so I don't have to break it into parts....? $\endgroup$ – akironjo Jan 18 '18 at 21:13
  • $\begingroup$ @akironjo Break the integral apart to a sum. Integration by parts or simple by-mind antiderivative tricks should do the job. Since I see you're a bit inexperienced regarding integration, check out the method "integration by parts" and apply it. $\endgroup$ – Rebellos Jan 18 '18 at 21:17

$$y''-6y'+9y=\frac{9x^2+6x+2}{x^3}$$ The solution of the associated homogeneous ODE $\quad y_h''-6y_h'+9y_h=0\quad$ is : $$\begin{cases}y_1=e^{3x}\\y_2=xe^{3x}\end{cases}\quad\to\quad y_h=c_1e^{3x}+c_2 xe^{3x}$$ The Wronskian is : $\quad W=\left|\begin{matrix}e^{3x} & xe^{3x} \\ 3e^{3x} & (3x+1)e^{3x} \end{matrix}\right|=e^{6x}$

With $\quad r=\frac{9x^2+6x+2}{x^3}$

$$y_p=-y_1\int \frac{y_2r}{W}dx+y_2\int \frac{y_1r}{W}dx$$

$$y_p=-e^{3x}\int \frac{xe^{3x} \frac{9x^2+6x+2}{x^3}}{e^{6x}}dx+xe^{3x}\int \frac{e^{3x} \frac{9x^2+6x+2}{x^3}}{e^{6x}}dx$$ $$y_p=-e^{3x}\left(-e^{-3x}\left(\frac2x+3\right)\right)+xe^{3x}\left(e^{-3x}\left(-\frac{1}{x^2}-\frac3x\right)\right)$$ $$y_p=\frac1x$$ $$y=c_1e^{3x}+c_2 xe^{3x}+\frac1x$$

  • $\begingroup$ Thank you very much! Could you explain me me how did you get from this part (that is my problem) yp=−e3x∫xe3x9x2+6x+2x3e6xdx+xe3x∫e3x9x2+6x+2x3e6xdx to this one yp=−e3x(−e−3x(2x+3))+xe3x(e−3x(−1x2−3x)) $\endgroup$ – akironjo Jan 21 '18 at 1:20
  • $\begingroup$ $\int e^{-3x}(9+\frac6x+\frac2{x^2})dx=e^{-3x}(-3-\frac2x)$ and $\int e^{-3x}(\frac9x+\frac6{x^2}+\frac2{x^3})dx=e^{-3x}(-\frac3x-\frac1{x^2})$ $\endgroup$ – JJacquelin Jan 21 '18 at 8:34
  • $\begingroup$ If you don't succeed with integration by part, use series : $\int e^{-3x}(9+\frac6x+\frac2{x^2})dx= e^{-3x}\left(a+\frac{b}{x}+\frac{c}{x^2}+\frac{d}{x^3}+... \right)$. Derivation and identification leads to $a=-3\:;\:b=-2\:;\:c=0\:;\:d=0\:;\:...$. Same method for the second integral. $\endgroup$ – JJacquelin Jan 21 '18 at 8:55

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