Consider $$ - u''(t) + p(t) u'(t) + q(t) u(t) = f(t), \quad t \in (a,b). $$ For $s \in (a,b)$ find coefficient functions $c_1$ and $c_2$ from the general solution of the homogenous equation $$ u_{\text{hom}}(t) = c_1 u_1(t) + c_2(t) $$ such that $u_{\text{hom}}(s) = 0$ and $u'_{\text{hom}}(s) = - f(s)$. Then $$ u_p(t) := \int_{t_0}^{t} c_1(s) u_1(t) + c_2(s) u_2(t) $$ is a particular solution of the inhomogeneous equation.

And now I am supposed to find a particular solution to $$ t^2(1 - t) u''(t) + 2t (2 - t) u'(t) + 2(1 + t) u(t) = \frac{1}{t - 1}, \quad t \in (0,1). $$

I have found the solution for the homogeneous equation to be $$ u_{\text{hom}}(t) = c_1 t^{-2} + c_2 \cdot \frac{t^2 - 3t + 3}{3t} $$ But the instructions confuse me in particular because in one equation $c_i$ seem to be constants and the other functions and I con't know how they are connected or if that is the case at all. Can somebody please give me a hint on how to continue?

Also, in the first equation of the yellow block, the highest derivate doesn't have a coefficient function, but since $p$ and $q$ are not relevant to the instructions that follow I don't have to divide my equation by $t^2(t - 1)$, right?

So taking the hints from the answer below I got: So my "Wronski-Matrix" is $$\begin{pmatrix} t^{-2} & \frac{t^2 -3t + 3}{3t} \\ -2 t^{-3} & \frac{1}{3} - t^{-2}\end{pmatrix},$$ solving I get $$C_1'(s) = \frac{t^3(t^2 - 3t + 3)}{(t - 3)^2(t - 1)}$$ and $$C_2'(t) = \frac{3t^2}{(1 - t)(t - 3)^2}.$$ Integrating those gives horrible terms, so what have I done wrong?

  • $\begingroup$ Otherwise known as "method of variation of constants". Cauchy wrote many nice textbooks and formulated first or almost first the standard theorems of modern calculus. Thus there is an inflationary occurrence of his name in this topic. Use alternate designations if they exist. $\endgroup$ – LutzL Aug 2 at 22:06

Do it literally as described: For any $s$ solve $u_{\rm hom}(s,c_1,c_2)=u_{\rm hom}'(s,c_1,c_2)=0$ and use these coefficients for this $s$ to set the values in the "variating constant coefficient functions" $C_1'(s)=c_1$ and $C_2'(s)=c_2$. For a different $s$ you have a different system to solve, thus get different constants.

Or in a less mystified way, solve \begin{align} \pmatrix{u_1(s)&u_2(s)\\u_1'(s)&u_2'(s)}\pmatrix{C_1'(s)\\C_2'(s)} =\pmatrix{0\\-f(s)} \end{align} and then integrate $C_1',C_2'$. This is commonly known an the method of variation of constants.

For that method to work you need to bring the equation into the normal form given in the theorem, so $$f(t)=\frac1{t^2(1-t)^2}.$$ This gives less horrible terms $$ C_1(t) = - \frac{\log{(1 - t)}}{3} + \frac{1}{9 (t-1)^{3} }, \quad C_2(t) = - \frac{1}{3 (t-1)^{3}} $$ that also partially cancel in the complete solution $$ y_p(t) = \frac{3 \log{(1 - t)} + 1}{9 t^{2}} $$

  • $\begingroup$ $u'_{\text{hom}}(s,c_1, c_2) = 0$? I thought $-f(s)$? $\endgroup$ – Viktor Glombik Aug 3 at 2:55
  • $\begingroup$ If you apply the theorem you have to transform your equation to the form given in the theorem. Added section about what results from that correction. $\endgroup$ – LutzL Aug 7 at 19:58

Given the homogeneous solution for

$$ t^2(1 - t) u''(t) + 2t (2 - t) u'(t) + 2(1 + t) u(t) = \frac{1}{t - 1}, \quad t \in (0,1). $$


$$ u_h = c_1 t^{-2} + c_2 \cdot \frac{t^2 - 3t + 3}{3t} $$

and considering $c_i = c_i(t)$ after substitution into the full DE we have

$$ 3 (t-1)^2 c_1''(t)-6 (t-1) c_1'(t)+t ((t-3) t+3) (t-1)^2 c_2''(t)+2 (2 t ((t-3) t+3)-3) (t-1) c_2'(t)+3= 0 $$

now solving for

$$ 3 (t-1)^2 c_1''(t)-6 (t-1) c_1'(t) = 0\\ t ((t-3) t+3) (t-1)^2 c_2''(t)+2 (2 t ((t-3) t+3)-3) (t-1) c_2'(t)+3= 0 $$

or after simplifications

$$ 3 (t-1) c_1''(t)-6 c_1'(t) = 0\\ t ((t-3) t+3) (t-1) c_2''(t)+2 (2 t ((t-3) t+3)-3) c_2'(t)+\frac{3}{t-1}= 0 $$

we obtain the $c_1(t), c_2(t)$. As we can see the $c_2(t)$ DE is also complicated. We could also consider as well the set of DEs

$$ 3 (t-1) c_1''(t)-6 c_1'(t) +\frac{3}{t-1} = 0\\ t ((t-3) t+3) (t-1) c_2''(t)+2 (2 t ((t-3) t+3)-3) c_2'(t)= 0 $$

perhaps a little simpler.


The variation of constants method (Lagrange) give us to solve DEs a degree lower than the original. This normally is worth a lot.


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.