2007 H2 Mathematics Paper 1 Question 4

Differential Equations (DEs)

Answers

I=2+4e−34t3{I = \frac{2+4\mathrm{e}^{- \frac{3}{4}t}}{3}}
The current approaches 23{\frac{2}{3}} for large values of t{t }

Full solutions

4dIdt=2−3I42−3IdIdt=1∫42−3I  dI=∫1  dt−43ln⁡∣2−3I∣=t+cln⁡∣2−3I∣=−34t−34c∣2−3I∣=e−34t−34c∣2−3I∣=e−34ce−34t2−3I=Ae−34t\begin{gather*} 4 \frac{\mathrm{d}I}{\mathrm{d}t} = 2 - 3 I \\ \frac{4}{2 - 3 I} \frac{\mathrm{d}I}{\mathrm{d}t} = 1 \\ \int \frac{4}{2 - 3 I} \; \mathrm{d}I = \int 1 \; \mathrm{d}t \\ - \frac{4}{3} \ln \left| 2 - 3 I\right| = t + c \\ \ln | 2-3I | = - \frac{3}{4}t - \frac{3}{4}c \\ | 2-3I | = \mathrm{e}^{- \frac{3}{4}t - \frac{3}{4}c} \\ | 2-3I | = \mathrm{e}^{- \frac{3}{4}c}\mathrm{e}^{- \frac{3}{4}t} \\ 2 - 3I = A \mathrm{e}^{- \frac{3}{4}t} \end{gather*}
When t=0,{t=0, } I=2{I = 2}
2−3(2)=Ae−34(0)A=−4\begin{align*} 2 - 3(2) &= A \mathrm{e}^{- \frac{3}{4}(0)} \\ A &= -4 \end{align*}
2−3I=−4e−34tI=2+4e−34t3  ■\begin{gather*} 2 - 3I = -4 \mathrm{e}^{- \frac{3}{4}t} \\ I = \frac{2+4\mathrm{e}^{- \frac{3}{4}t}}{3} \; \blacksquare \end{gather*}
As t→∞,{t \to \infty, } e−34t→0.{\mathrm{e}^{- \frac{3}{4}t} \to 0.} Hence θ→23{\theta \to \frac{2}{3}}
The current approaches 23{\frac{2}{3}} for large values of t  ■{t \; \blacksquare}