differential equation

450 DIFFERENTIAL EQUATIONS
Problem 13. For an adiabatic expansion of 2. (2y − 1) dy
agas
dx = (3x2 + 1), given x = 1 when
y = 2. [y 2 − y = x 3 + x]
dp
C v
p + C dV
p
V = 0,
dy
3.
dx = e2x−y ,givenx = 0 when y = 0.
where C p and C v are constants. Given n = C [
p
,
e
C y = 1
v
show that pV n 2 e2x + 1 ]
2
= constant.
4. 2y(1 − x) + x(1 + y) dy = 0, given x = 1
dx
Separating the variables gives:
when y = 1. [ln (x 2 y) = 2x − y − 1]
dp
C v
p =−C dV
5. Show that the solution of the equation
p
V
y 2 + 1
x 2 + 1 = y dy
is of the form
x dx
Integrating both sides gives:
√ (y 2 )
+ 1
∫ ∫ dp dV x 2 = constant.
+ 1
C v
p =−C p
V
6. Solve xy = (1 − x 2 ) dy for y, given x = 0
dx [
]
i.e. C v ln p =−C p ln V + k
1
when y = 1.
y = √
(1 − x 2 )
Dividing throughout by constant C v gives:
ln p =− C p
ln V + k 7. Determine the equation of the curve which
satisfies the equation xy dy
C v C v
dx = x2 − 1, and
which passes through the point (1, 2).
Since C p
= n, then ln p + n ln V = K,
[y 2 = x 2 − 2lnx + 3]
C v
where K = k 8. The p.d., V, between the plates of a capacitor
C charged by a steady voltage E
.
C v through a resistor R is given by the equation
i.e. ln p + ln V n = K or ln pV n = K, by the laws of CR dV
logarithms.
dt + V = E.
Hence pV n = e K , i.e., pV n (a) Solve the equation for V given that at
= constant.
t = 0, V = 0.
(b) Calculate V, correct to 3 significant figures,
when E = 25 V, C = 20 ×10 −6 F,
Now try the following exercise.
R = 200 ×10 3 and t = 3.0s.
⎡
( ) ⎤
Exercise 180 Further problems on equations
of the form dy
⎣ (a) V = E 1 − e CR
−t
⎦
= f (x) · f (y)
dx (b) 13.2V
In Problems 1 to 4, solve the differential 9. Determine the value of p, given that
equations.
x 3 dy = p − x, and that y = 0 when x = 2 and
dy
1.
dx = 2y cos x [ln y = 2 sin x + c] dx
when x = 6. [3]