Solve ##\int\frac{e^{-x}}{x^2}dx## and ##\int \frac{e^{-x}}{x}dx##

[zenterix]

Homework Statement

Find the general solution to the differential equation

Relevant Equations

$y''+2y'+y=\frac{e^{-x}}{x^2}$

The characteristic equation has a zero discriminant and the sole root of $-1$.

The general solution to the associated homogeneous equation is thus

$$y_h(x)=e^{-x}(c_1+c_2x)\tag{1}$$

Now we only need to find one particular solution of the non-homogeneous equation.

The righthand side of the non-homogeneous equation, call it $R(x)$, is not a polynomial, nor does it fall in the category of $p(x)e^{mx}$ with $p$ being a polynomial. These are perhaps the two easiest special cases of $R(x)$ to solve because they can be solved by the method of undetermined coefficients.

$R(x)$ is also not an exponential multiplied by an expression containing $\sin$ and/or $\cos$, which can also be solved by the method of undetermined coefficients.

Thus, I have chosen to solve using a more general method based on the following theorem in Apostol's Calculus, Volume I

We have $v_1(x)=1$ and $v_2(x)=x$ as our solutions to $L(y)=0$, ie the homogeneous equation.

The Wronskian of $v_1$ and $v_2$ is

$$W(x)=
\begin{vmatrix}
1 & 0\\
x & 1
\end{vmatrix}=1
\tag{2}$$

Thus,

$$t_1(x)=-\int x\cdot\frac{e^{-x}}{x^2}dx=\int\frac{e^{-x}}{x}dx\tag{3}$$

$$t_2(x)=\int \frac{e^{-x}}{x^2}dx\tag{4}$$

How do we solve these integrals?

Math software doesn't get me very far:

The window on the bottom is a sort of "math tutor" in Maple that shows all the steps in the calculation of the integral. However, for some integrals this step-by-step isn't available.

Now $i_1(x)$ is probably some variable defined within the depths of the dungeons of Maple documentation.

Still, what technique for solving integrals works for these integrals?

 

[anuttarasammyak]

The integral is not expresed by elementaly functions.

See https://en.wikipedia.org/wiki/Exponential_integral .

 

[zenterix]

In that case, here is a follow up question.

The solution to the problem in the back of the book is

$$y=(c_1+c_2x-\log{|x|})e^{-x}$$

which seems to imply that the particular solution they are using for the non-homogeneous equation is

$$y_p(x)=-e^{-x}\log{|x|}$$

How did they arrive at this particular solution?

 

[PeroK]

$$y_p(x)=-e^{-x}\log{|x|}$$

How did they arrive at this particular solution?

I would look for a function of the form $y_p(x) = e^{-x}f(x)$. Even without seeing the solution, that looks like a good idea.

 

[anuttarasammyak]

Substition
y=z e^{-x}
seems helpful. Have you tried it ?

 

[zenterix]

I would look for a function of the form $y_p(x) = e^{-x}f(x)$. Even without seeing the solution, that looks like a good idea.

Indeed it is a good idea and the answer comes out quite easily.

 

[pasmith]

Homework Statement: Find the general solution to the differential equation
Relevant Equations: $y''+2y'+y=\frac{e^{-x}}{x^2}$

Consider the identity <br /> \frac{d^2}{dx^2} (e^{ax}g) = (g&#039;&#039; + 2ag&#039; + a^2g)e^{ax}.

 

[TSny]

The general solution to the associated homogeneous equation is thus

$$y_h(x)=e^{-x}(c_1+c_2x)\tag{1}$$

We have $v_1(x)=1$ and $v_2(x)=x$ as our solutions to $L(y)=0$, ie the homogeneous equation.

Shouldn't that be $v_1(x) =e^{-x}$ and $v_2(x) = x e^{-x}$?

The method of solution given by the theorem in Apostol will work out nicely. [Edited to add this comment.]