Réitigh y=E(1-c^-t/4)

$$y = E ( 1 - c ^ { - t / 4 } )$$

Réitigh do E. (complex solution)

$$\left\{\begin{matrix}E=-\frac{yc^{\frac{t}{4}}}{1-c^{\frac{t}{4}}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }c=e^{-\frac{\pi n_{1}iRe(t)}{2\times \frac{\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}}{16}}-\frac{\pi n_{1}Im(t)}{2\times \frac{\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}}{16}}}\\E\in \mathrm{C}\text{, }&\left(c=0\text{ and }t\neq 0\right)\text{ or }\left(y=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }c=e^{-\frac{\pi n_{1}iRe(t)}{2\times \frac{\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}}{16}}-\frac{\pi n_{1}Im(t)}{2\times \frac{\left(Re(t)\right)^{2}+\left(Im(t)\right)^{2}}{16}}}\right)\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun E a mhéadú faoi 1-c^{\frac{-t}{4}}.

$$y=E-Ec^{\frac{-t}{4}}$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$E-Ec^{\frac{-t}{4}}=y$$

Athordaigh na téarmaí.

$$-Ec^{-\frac{t}{4}}+E=y$$

Comhcheangail na téarmaí ar fad ina bhfuil E.

$$\left(-c^{-\frac{t}{4}}+1\right)E=y$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(1-c^{-\frac{t}{4}}\right)E=y$$

Roinn an dá thaobh faoi -c^{-\frac{1}{4}t}+1.

$$\frac{\left(1-c^{-\frac{t}{4}}\right)E}{1-c^{-\frac{t}{4}}}=\frac{y}{1-c^{-\frac{t}{4}}}$$

Má roinntear é faoi -c^{-\frac{1}{4}t}+1 cuirtear an iolrúchán faoi -c^{-\frac{1}{4}t}+1 ar ceal.

$$E=\frac{y}{1-c^{-\frac{t}{4}}}$$

Roinn y faoi -c^{-\frac{1}{4}t}+1.

$$E=\frac{yc^{\frac{t}{4}}}{c^{\frac{t}{4}}-1}$$