Chapter 9 Differential Equations Ex 9.6 Solutions
Question - 11 : - 
Answer - 11 : -
This is a linear differential equation of the form:
The general solution of the given differential equation isgiven by the relation,
Answer - 12 : -
This is a linear differential equation of the form:
The general solution of the given differential equation isgiven by the relation,
Answer - 13 : -
The given differential equation is 
This is a linear equation of the form:
The general solution of the given differential equation isgiven by the relation,
Now,
Therefore,
Substituting C = –2 in equation (1), we get:
Hence, the required solution of the givendifferential equation is 
Question - 14 : - 
Answer - 14 : -
This is a linear differential equation of the form:
The general solution of the given differential equation isgiven by the relation,
Now, y = 0 at x =1.
Therefore,
Substituting 
in equation (1), we get:
This is the required general solution of thegiven differential equation.
Answer - 15 : -
The given differential equation is 
This is a linear differential equation of the form:
The general solution of the given differential equation isgiven by the relation,
Now,
Therefore, we get:
Substituting C = 4 in equation (1), we get:
This is the required particular solution ofthe given differential equation.
Question - 16 : - Find the equation of a curve passing throughthe origin given that the slope of the tangent to the curve at any point (x, y)is equal to the sum of the coordinates of the point.
Answer - 16 : -
Let F (x, y)be the curve passing through the origin.
At point (x, y), theslope of the curve will be
According to the given information:
This is a linear differential equation of the form:
The general solution of the given differential equation isgiven by the relation,
Substituting in equation (1), we get:
The curve passes through the origin.
Therefore, equation (2) becomes:
1 = C
⇒ C = 1
Substituting C = 1 in equation (2), we get:
Hence, the required equation of curvepassing through the origin is
Find the equation of a curve passing throughthe point (0, 2) given that the sum of the coordinates of any point on the curveexceeds the magnitude of the slope of the tangent to the curve at that point by5.
Answer - 17 : -
Let F (x, y)be the curve and let (x, y) be a point on the curve. Theslope of the tangent to the curve at (x, y) is
According to the given information:
This is a linear differential equation of the form:
The general equation of the curve is given by therelation,
Therefore, equation (1) becomes:
The curve passes through point (0, 2).
Therefore, equation (2) becomes:
0 + 2 – 4 = Ce0
⇒ – 2 = C
⇒ C = – 2
Substituting C = –2 in equation (2), we get:
This is the required equation of the curve.
Question - 18 : - The integrating factor of the differential equation
is
A. e–x
B. e–y
C. 
D. x
Answer - 18 : -
The given differential equation is:
This is a linear differential equation of the form:
The integrating factor (I.F) is given by the relation,
Hence, the correct answer is C.
Question - 19 : - The integrating factor of the differential equation.
is
A.
B.
C.
D.
Answer - 19 : -
The given differential equation is:
This is a linear differential equation of the form:
The integrating factor (I.F) is given by the relation,
Hence, the correct answer is D.