Slučaj $\boldsymbol {q=0.}$ (nema otpora sredstva)

Jednadžba je tada

$$\left(p(x)\,u'(x)\right)' + f(x) = 0,$$

pa imamo

$$\left(p(x)\,u'(x)\right)' = - f(x) \hspace{1cm} \left/\int_0^x d\xi,\right.$$

$$p(x)\,u'(x) -p(0)\,u'(0) = -\int_0^x f(\xi)\,d\xi,$$

$$u'(x)= \frac{p(0)\,u'(0)}{p(x)} - \frac{1}{p(x)}\int_0^x f(\xi)\,d\xi \hspace{1cm}\left/\int_0^x d\eta,\right.$$ (2.10)

$$u(x)-u(0)= p(0)\,u'(0)\,\int_0^x \frac{d\eta}{p(\eta)} - \int_0^x \left(\frac{1}{p({\eta})}\int_0^{\eta} f(\xi)\,d\xi\right) \,d\eta,$$

tj.

$$u(x)=u(0)+ p(0)\,u'(0)\,\int_0^x \frac{d\eta}{p(\eta)} - \int_0^x \left(\frac{1}{p({\eta})}\int_0^{\eta} f(\xi)\,d\xi\right) \,d\eta.$$ (2.11)

U ovoj formuli imamo dvije neodređene veličine $u(0)$ i $p(0)u'(0).$ Razmotrimo sada jedinstvenost uz navedene rubne uvjete.