1. **Stating the problem:** We have an electric circuit with two loops and currents $i_1(t)$ and $i_2(t)$. We need to: a. Formulate the system of differential equations for $i_1(t)$ and $i_2(t)$. b. Solve the system using the variation of parameters method given: $R_1=8$, $R_2=3$, $L_1=1$, $L_2=1$, $E=100\sin t$, and initial conditions $i_1(0)=0$, $i_2(0)=0$. 2. **Formulating the system of differential equations:** Using Kirchhoff's voltage law (KVL) for each loop: - Loop 1 (with $i_1$): $$L_1 \frac{di_1}{dt} + R_1 i_1 + L_2 \frac{di_2}{dt} = E(t)$$ - Loop 2 (with $i_2$): $$L_2 \frac{di_1}{dt} + R_2 i_2 + L_2 \frac{di_2}{dt} = 0$$ Since $L_2$ appears in both loops, the system can be written as: $$\begin{cases} L_1 \frac{di_1}{dt} + R_1 i_1 + L_2 \frac{di_2}{dt} = E(t) \\ L_2 \frac{di_1}{dt} + R_2 i_2 + L_2 \frac{di_2}{dt} = 0 \end{cases}$$ 3. **Substitute given values:** $$\begin{cases} 1 \cdot \frac{di_1}{dt} + 8 i_1 + 1 \cdot \frac{di_2}{dt} = 100 \sin t \\ 1 \cdot \frac{di_1}{dt} + 3 i_2 + 1 \cdot \frac{di_2}{dt} = 0 \end{cases}$$ 4. **Rewrite system:** $$\begin{cases} \frac{di_1}{dt} + 8 i_1 + \frac{di_2}{dt} = 100 \sin t \\ \frac{di_1}{dt} + 3 i_2 + \frac{di_2}{dt} = 0 \end{cases}$$ 5. **Express derivatives:** Subtract second equation from first: $$\left(\frac{di_1}{dt} + 8 i_1 + \frac{di_2}{dt}\right) - \left(\frac{di_1}{dt} + 3 i_2 + \frac{di_2}{dt}\right) = 100 \sin t - 0$$ Simplifies to: $$8 i_1 - 3 i_2 = 100 \sin t$$ 6. **Solve for $i_2$ in terms of $i_1$:** $$3 i_2 = 8 i_1 - 100 \sin t \implies i_2 = \frac{8}{3} i_1 - \frac{100}{3} \sin t$$ 7. **Substitute $i_2$ into second original equation:** Second equation: $$\frac{di_1}{dt} + 3 i_2 + \frac{di_2}{dt} = 0$$ Calculate $\frac{di_2}{dt}$: $$\frac{di_2}{dt} = \frac{8}{3} \frac{di_1}{dt} - \frac{100}{3} \cos t$$ Substitute: $$\frac{di_1}{dt} + 3 \left(\frac{8}{3} i_1 - \frac{100}{3} \sin t\right) + \frac{8}{3} \frac{di_1}{dt} - \frac{100}{3} \cos t = 0$$ Simplify: $$\frac{di_1}{dt} + 8 i_1 - 100 \sin t + \frac{8}{3} \frac{di_1}{dt} - \frac{100}{3} \cos t = 0$$ Combine derivatives: $$\left(1 + \frac{8}{3}\right) \frac{di_1}{dt} + 8 i_1 = 100 \sin t + \frac{100}{3} \cos t$$ $$\frac{11}{3} \frac{di_1}{dt} + 8 i_1 = 100 \sin t + \frac{100}{3} \cos t$$ 8. **Rewrite as first order ODE for $i_1$:** $$\frac{di_1}{dt} + \frac{24}{11} i_1 = \frac{300}{11} \sin t + \frac{100}{11} \cos t$$ 9. **Solve the ODE using variation of parameters:** - Homogeneous equation: $$\frac{di_1}{dt} + \frac{24}{11} i_1 = 0$$ Solution: $$i_{1h} = C e^{-\frac{24}{11} t}$$ - Particular solution $i_{1p}$: Assume: $$i_{1p} = A \sin t + B \cos t$$ Derive: $$\frac{di_{1p}}{dt} = A \cos t - B \sin t$$ Substitute into ODE: $$A \cos t - B \sin t + \frac{24}{11} (A \sin t + B \cos t) = \frac{300}{11} \sin t + \frac{100}{11} \cos t$$ Group terms: $$\left(A + \frac{24}{11} B\right) \cos t + \left(-B + \frac{24}{11} A\right) \sin t = \frac{100}{11} \cos t + \frac{300}{11} \sin t$$ Equate coefficients: $$A + \frac{24}{11} B = \frac{100}{11}$$ $$-B + \frac{24}{11} A = \frac{300}{11}$$ 10. **Solve system for $A$ and $B$:** Multiply first by 11: $$11 A + 24 B = 100$$ Multiply second by 11: $$-11 B + 24 A = 300$$ Rewrite: $$11 A + 24 B = 100$$ $$24 A - 11 B = 300$$ Multiply first by 11 and second by 24 to eliminate $B$: $$121 A + 264 B = 1100$$ $$576 A - 264 B = 7200$$ Add: $$697 A = 8300 \implies A = \frac{8300}{697}$$ Substitute $A$ into first: $$11 \cdot \frac{8300}{697} + 24 B = 100$$ $$\frac{91300}{697} + 24 B = 100$$ $$24 B = 100 - \frac{91300}{697} = \frac{69700 - 91300}{697} = -\frac{21600}{697}$$ $$B = -\frac{21600}{697 \cdot 24} = -\frac{21600}{16728} = -\frac{900}{697}$$ 11. **Final particular solution:** $$i_{1p} = \frac{8300}{697} \sin t - \frac{900}{697} \cos t$$ 12. **General solution for $i_1$:** $$i_1(t) = C e^{-\frac{24}{11} t} + \frac{8300}{697} \sin t - \frac{900}{697} \cos t$$ 13. **Apply initial condition $i_1(0) = 0$:** $$0 = C + 0 - \frac{900}{697} \implies C = \frac{900}{697}$$ 14. **Final solution for $i_1(t)$:** $$i_1(t) = \frac{900}{697} e^{-\frac{24}{11} t} + \frac{8300}{697} \sin t - \frac{900}{697} \cos t$$ 15. **Find $i_2(t)$ using step 6:** $$i_2 = \frac{8}{3} i_1 - \frac{100}{3} \sin t$$ Substitute $i_1(t)$: $$i_2(t) = \frac{8}{3} \left(\frac{900}{697} e^{-\frac{24}{11} t} + \frac{8300}{697} \sin t - \frac{900}{697} \cos t\right) - \frac{100}{3} \sin t$$ Simplify: $$i_2(t) = \frac{7200}{2091} e^{-\frac{24}{11} t} + \left(\frac{66400}{2091} - \frac{700}{21}\right) \sin t - \frac{7200}{2091} \cos t$$ **Summary:** - System of differential equations derived. - Solved $i_1(t)$ explicitly. - Expressed $i_2(t)$ in terms of $i_1(t)$. **Final answers:** $$i_1(t) = \frac{900}{697} e^{-\frac{24}{11} t} + \frac{8300}{697} \sin t - \frac{900}{697} \cos t$$ $$i_2(t) = \frac{7200}{2091} e^{-\frac{24}{11} t} + \left(\frac{66400}{2091} - \frac{700}{21}\right) \sin t - \frac{7200}{2091} \cos t$$