Description

Mathematica
Problem Sheet 6
dy
1. Solve the differential equation  sin5x dx
2. a) Solve the equation for y as a function of x
2 dy y  yx
x  dx
b) Find a special solution of the above mentioned equation for the following initial value
1 y(1) 
e
3. a) Solve the equation for y as a function of x d2y dy
dx2 5 dx  2y  0
b) Find a special solution of the above mentioned equation for the following conditions y(0)  5, y ‘ (0) 10
4. Solve the following equation numerically for y, for the given conditions and the given interval of x
d2y
2 xy0 dx
y(0) 1, y(1) 1
0x10
5. Solve the following equation numerically for y, for the given conditions and the given interval of
t
d2y dy 12 dy  y 0 dt2  dt  dt

y(0)  1, y'(0)  0
0t 10
6. Solve the following equation numerically for y, for the given conditions and the given interval of x
d2y dy
2  0.3dx sin y  0
dx
y(0)  2, y'(0)  0
0 x 30

7. See the following diagram of an RLC circuit where V is the AC voltage source

Here, R = resistance of the resistor
C = capacitance of the capacitor L = inductance of the inductor i = current flow through the circuit From Electromagnetism we know
dQ
Voltage across the resistor, VR = i R R . Here Q is charge and t is time dt
Q
Voltage across the capacitor, VC = where Q is the charged stored in the capacitor.
C
di d2Q
Voltage across the inductor, VL  L dt  L dt2
The electromotive force is given by
V V0eit
Applying Kirchoff’s loop voltage law we can write
VVR VC VL
d 2Q dQ 1 jt
 L 2  R dt  C Q V0e dt

Now solve this equation for Q as a function of t
Then find i as a function of t by just differentiating Q with respect to t