Chapter 9 Differential Equations Ex 9.4 Solutions
Question - 11 : - 
Answer - 11 : -
Thegiven differential equation is:
Integratingboth sides, we get:
Comparing the coefficients of x2 and x,we get:
A + B = 2
B + C = 1
A + C = 0
Solvingthese equations, we get:
Substitutingthe values of A, B, and C in equation (2), we get:
Therefore,equation (1) becomes:
SubstitutingC = 1 in equation (3), we get:
Question - 12 : - 
Answer - 12 : -
Integratingboth sides, we get:
Comparing the coefficients of x2, x, andconstant, we get:
Solvingthese equations, we get
Substituting the values of A, B, and C inequation (2), we get:
Therefore,equation (1) becomes:
Substituting the value of k2 in equation (3), we get:
Answer - 13 : -
Integratingboth sides, we get:
SubstitutingC = 1 in equation (1), we get:
Answer - 14 : -
Integratingboth sides, we get:
SubstitutingC = 1 in equation (1), we get:
y = sec x
Findthe equation of a curve passing through the point (0, 0) and whose differentialequation is
.
Answer - 15 : -
Thedifferential equation of the curve is:
Integratingboth sides, we get:
Substitutingthis value in equation (1), we get:
Now,the curve passes through point (0, 0).
Substituting 
in equation (2), we get:
Hence,the required equation of the curve is 
Question - 16 : - Forthe differential equation 
find the solution curve passingthrough the point (1, –1).
Answer - 16 : -
Thedifferential equation of the given curve is:
Integratingboth sides, we get:
Now,the curve passes through point (1, –1).
SubstitutingC = –2 in equation (1), we get:
Thisis the required solution of the given curve.
Question - 17 : - Find the equation of a curve passing throughthe point (0, –2) given that at any point (x, y) on the curve, the productof the slope of its tangent and y-coordinate of the point is equalto the x-coordinate of the point.
Answer - 17 : -
Let x and y bethe x-coordinate and y-coordinate of the curverespectively.
Weknow that the slope of a tangent to the curve in the coordinate axis is givenby the relation,
Accordingto the given information, we get:
Integratingboth sides, we get:
Now,the curve passes through point (0, –2).
∴ (–2)2 – 02 = 2C
⇒ 2C = 4
Substituting2C = 4 in equation (1), we get:
y2 – x2 = 4
Thisis the required equation of the curve.
Question - 18 : - At any point (x, y) of acurve, the slope of the tangent is twice the slope of the line segment joiningthe point of contact to the point (–4, –3). Find the equation of the curvegiven that it passes through (–2, 1).
Answer - 18 : -
It is given that (x, y)is the point of contact of the curve and its tangent.
The slope (m1) of the line segmentjoining (x, y) and (–4, –3) is 
Weknow that the slope of the tangent to the curve is given by the relation,
Accordingto the given information:
Integratingboth sides, we get:
Thisis the general equation of the curve.
Itis given that it passes through point (–2, 1).
SubstitutingC = 1 in equation (1), we get:
y + 3 = (x + 4)2
Thisis the required equation of the curve.
Question - 19 : - The volume of spherical balloon beinginflated changes at a constant rate. If initially its radius is 3 units andafter 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Answer - 19 : -
Let the rate of change of the volume of theballoon be k (where k is a constant).
Integratingboth sides, we get:
⇒ 4π × 33 = 3 (k × 0 +C)
⇒ 108π = 3C
⇒ C = 36π
At t = 3, r =6:
⇒ 4π × 63 = 3 (k ×3 + C)
⇒ 864π = 3 (3k +36π)
⇒ 3k = –288π – 36π= 252π
⇒ k = 84π
Substituting the values of k andC in equation (1), we get:
Thus,the radius of the balloon after t seconds is
.
Question - 20 : - In a bank, principal increases continuouslyat the rate of r% per year. Find the value of r ifRs 100 doubles itself in 10 years (loge 2 = 0.6931).
Answer - 20 : -
Let p, t, and r representthe principal, time, and rate of interest respectively.
It is given that the principal increasescontinuously at the rate of r% per year.
Integratingboth sides, we get:
It is given that when t =0, p = 100.
⇒ 100 = ek … (2)
Now, if t = 10, then p =2 × 100 = 200.
Therefore,equation (1) becomes:
Hence,the value of r is 6.93%.