Question: ZEAL EDUCATION SOCIETY’S ZEAL COLLEGE OF ENGINEERING AND RESEARCH NARHE | PUNE -41 | INDIA ZCOER-ACAD/R/16F Revision: 00 Date:01/04/2 QUESTION BANK Electronics and Computer Engineering Semester: III Academic Yea Tech Div: A,B Date: Engineering Mathematics – III Course Code: 25ECMD301 ewise from the attached 2025 Pattern syllabus. Unit 1 : Linear Differential Equations Q. No. Question 1 (D^2+2D + 1) y = 4sin2x 2 (D^2-9) y = e^3x+1 3 (D^2− 2D + 1) y = e^xsinx 4 (D^2 + 4) y = sec2x by method of variation of parameter. 5 (D^2 − 6D + 9) y = e^3x/x^2 by method of variation of parameters. 6 d^2y/dx^2 + y = Cosec x by method of variation of parameter. 7 dx/(y−z) = dy/(z−x) = dz/(x−y) 8 dx/x = dy/y = dz/z 9 x^2 d^2y/dx^2 − 4x dy/dx + 6y = x^5 10 (x + 1)^2 d^2y/dx^2 + (x + 1) dy/dx + y = 2 sin(log(x+1)) 11 An electric current consists of inductance 0.1 H, resistance R as 20 ohms and a condenser of of 25 micro farad. If the differential equatio circuit is L d^2q/dt^2 + R dq/dt + q/c = 0 Then find the charge q and current i at any time t , Given that t=0, q=0.05 coulombs, i=dq/dt=0, When t=0

1. **Problem Statement:** Find the charge $q(t)$ and current $i(t)$ at any time $t$ for an electric circuit with inductance $L=0.1$ H, resistance $R=20$ ohms, and capacitance $C=25$ microfarads, governed by the differential equation: $$L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = 0$$ with initial conditions $q(0) = 0.05$ coulombs and $i(0) = \frac{dq}{dt}(0) = 0$. 2. **Given:** - $L = 0.1$ - $R = 20$ - $C = 25$ microfarads $= 25 \times 10^{-6}$ F - Initial conditions: $q(0) = 0.05$, $i(0) = 0$ 3. **Rewrite the differential equation:** $$0.1 \frac{d^2 q}{dt^2} + 20 \frac{dq}{dt} + \frac{q}{25 \times 10^{-6}} = 0$$ Divide through by $0.1$: $$\frac{d^2 q}{dt^2} + 200 \frac{dq}{dt} + 400000 q = 0$$ 4. **Characteristic equation:** $$r^2 + 200 r + 400000 = 0$$ 5. **Solve the quadratic:** $$r = \frac{-200 \pm \sqrt{200^2 - 4 \times 1 \times 400000}}{2} = \frac{-200 \pm \sqrt{40000 - 1600000}}{2} = \frac{-200 \pm \sqrt{-1560000}}{2}$$ Since discriminant is negative, roots are complex: $$r = -100 \pm 100 \sqrt{39} i$$ 6. **General solution:** $$q(t) = e^{-100 t} (A \cos(100 \sqrt{39} t) + B \sin(100 \sqrt{39} t))$$ 7. **Apply initial conditions:** At $t=0$: $$q(0) = A = 0.05$$ Derivative: $$i(t) = \frac{dq}{dt} = e^{-100 t} \left(-100 A \cos(\omega t) - A \omega \sin(\omega t) - 100 B \sin(\omega t) + B \omega \cos(\omega t) \right)$$ where $\omega = 100 \sqrt{39}$. At $t=0$: $$i(0) = -100 A + B \omega = 0$$ Substitute $A=0.05$: $$-100 \times 0.05 + B \times 100 \sqrt{39} = 0 \Rightarrow -5 + 100 \sqrt{39} B = 0$$ Solve for $B$: $$B = \frac{5}{100 \sqrt{39}} = \frac{1}{20 \sqrt{39}}$$ 8. **Final solution:** $$q(t) = e^{-100 t} \left(0.05 \cos(100 \sqrt{39} t) + \frac{1}{20 \sqrt{39}} \sin(100 \sqrt{39} t) \right)$$ $$i(t) = \frac{dq}{dt}$$ as above. **Summary:** - Charge $q(t)$ decays exponentially with oscillations. - Current $i(t)$ is the time derivative of $q(t)$.