Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to get some help by solving the following problem:

$$ p'_1= \frac 1 x p_1 - p_2 + x$$ $$ p'_2=\frac 1{x^2}p_1+\frac 2 x p_2 - x^2 $$

with initial conditions $$p_1(1)=p_2(1)=0, x \gt0 $$


If I use Wolframalpha, I get

enter image description here

enter image description here

Where $u$ and $v$ are obviously $p_1$ and $p_2$. Can anybody explain whats going on?

share|cite|improve this question
up vote 3 down vote accepted

One approach is to express $p_2=-p_1'+\frac1xp_1+x$ from the first equation and substitute into the second to get: $$-p_1''+\frac1xp_1'-\frac1{x^2}p_1+1=p_2'=\frac1{x^2}p_1+\frac2x\left(-p_1'+\frac1xp_1+x\right)-x^2$$ $$p_1''-\frac3xp_1'+\frac4{x^2}p_1=x^2-1$$ Multiplying by $x^2$ we get $x^2p_1''-3xp_1'+4p_1=x^4-x^2$, which is a Cauchy–Euler equation.
Solve the homogeneous equation $x^2p_1''-3xp_1'+4p_1=0$ first: $p_1=x^r$, hence $$x^r(r(r-1)-3r+4)=0\hspace{5pt}\Rightarrow\hspace{5pt} r^2-4r+4=(r-2)^2=0 \hspace{5pt}\Rightarrow\hspace{5pt} r=2$$ So the solution to the homogeneous equation is $C_1x^2\ln x+C_2x^2$.
Now we can use Green's function of the equation to find $p_1$: ($y_1(x)=x^2\ln x,\hspace{3pt} y_2(x)=x^2$) $$\begin{align*}k(x,t)&=\frac{y_1(t)y_2(x)-y_1(x)y_2(t)}{y_1(t)y_2'(t)-y_2(t)y_1'(t)}= \frac{t^2\ln t\cdot x^2-x^2\ln x\cdot t^2}{t^2\ln t\cdot 2t-t^2\cdot(2t\ln t+t)}=\frac{t^2\ln t\cdot x^2-x^2\ln x\cdot t^2}{-t^3}\\ &=\frac{x^2\ln x-x^2\ln t}{t}\end{align*}$$ Then ($b(x)$ is the in-homogeneous part, i.e. $b(x)=x^4-x^2$) $$\begin{align*}p_1(x)&=\int k(x,t)b(t)dt=\int \frac{x^2\ln x-x^2\ln t}{t}t^2(t^2-1)dt\\ &=x^2\ln x\int t(t^2-1)dt-x^2\int t(t^2-1)\ln tdt\end{align*}$$ Compute the integral, find $p_1$ using you initial values and the substitute back to find $p_2$.

share|cite|improve this answer
Hmm I am told that I can use that a fundamental matrix of the homogeneous system has entries is given by the 2x2 matrix $\Psi$ with the following entries: $\Psi_{11} = x^2$; $\Psi_{12} = -x^2 \ln x$; $\Psi_{21} = -x $; $\Psi_{22} = x + x \ln x$ – Applied mathematician Jan 8 '13 at 16:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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