Differential Equations Practice Questions - DPP for JEE

 Page 2


1. (a)
;
or or    ( Q )
Squaring, 
Differentiating, 
(Here, y
2
 = ) or (1+x
2
)y
2
 + xy
1
 = n
2
y
2. (c) Divide the equation by 
Put 
Thus, we get,  , linear in z
? The solution is, 
Page 3


1. (a)
;
or or    ( Q )
Squaring, 
Differentiating, 
(Here, y
2
 = ) or (1+x
2
)y
2
 + xy
1
 = n
2
y
2. (c) Divide the equation by 
Put 
Thus, we get,  , linear in z
? The solution is, 
 
3. (a) Taking x = r cos ? and y = r sin ?, so that
x
2
 + y
2
 = r
2
 and 
we have x dx + y dy = r dr
and x dy – y dx = x
2
 sec
2
 ?d? = r
2
 d?.
The given equation can be transformed into
? ? 
Integrating both sides, then we get
? ...(i)
? sin {tan
–1
(y/x) + c}
? sin {tan
–1
 (y/x) + constant}
4. (c) Any conic whose axes coincide with co-ordinate axis is ax
2 
+ by
2 
=
1 ..(i)
Diff. both sides w.r.t. 'x', we get
2ax + 2by = 0  i.e. ax + by  = 0 ..(ii)
Diff. again,  a + b = 0 ..(iii)
From (ii),  
Page 4


1. (a)
;
or or    ( Q )
Squaring, 
Differentiating, 
(Here, y
2
 = ) or (1+x
2
)y
2
 + xy
1
 = n
2
y
2. (c) Divide the equation by 
Put 
Thus, we get,  , linear in z
? The solution is, 
 
3. (a) Taking x = r cos ? and y = r sin ?, so that
x
2
 + y
2
 = r
2
 and 
we have x dx + y dy = r dr
and x dy – y dx = x
2
 sec
2
 ?d? = r
2
 d?.
The given equation can be transformed into
? ? 
Integrating both sides, then we get
? ...(i)
? sin {tan
–1
(y/x) + c}
? sin {tan
–1
 (y/x) + constant}
4. (c) Any conic whose axes coincide with co-ordinate axis is ax
2 
+ by
2 
=
1 ..(i)
Diff. both sides w.r.t. 'x', we get
2ax + 2by = 0  i.e. ax + by  = 0 ..(ii)
Diff. again,  a + b = 0 ..(iii)
From (ii),  
From (iii),  
?
? = 0
5. (d) Putting v = y / x so that 
We have  
?
? log | C x | = 
(C being constant of integration)
But y =   is the general solution,
So 
? f (1/v) = – 1/v
2
(differentiating w.r.t. v both sides)
? f (x/y) = – y
2
 / x
2
6. (a) We have 
Divide by y
2
Page 5


1. (a)
;
or or    ( Q )
Squaring, 
Differentiating, 
(Here, y
2
 = ) or (1+x
2
)y
2
 + xy
1
 = n
2
y
2. (c) Divide the equation by 
Put 
Thus, we get,  , linear in z
? The solution is, 
 
3. (a) Taking x = r cos ? and y = r sin ?, so that
x
2
 + y
2
 = r
2
 and 
we have x dx + y dy = r dr
and x dy – y dx = x
2
 sec
2
 ?d? = r
2
 d?.
The given equation can be transformed into
? ? 
Integrating both sides, then we get
? ...(i)
? sin {tan
–1
(y/x) + c}
? sin {tan
–1
 (y/x) + constant}
4. (c) Any conic whose axes coincide with co-ordinate axis is ax
2 
+ by
2 
=
1 ..(i)
Diff. both sides w.r.t. 'x', we get
2ax + 2by = 0  i.e. ax + by  = 0 ..(ii)
Diff. again,  a + b = 0 ..(iii)
From (ii),  
From (iii),  
?
? = 0
5. (d) Putting v = y / x so that 
We have  
?
? log | C x | = 
(C being constant of integration)
But y =   is the general solution,
So 
? f (1/v) = – 1/v
2
(differentiating w.r.t. v both sides)
? f (x/y) = – y
2
 / x
2
6. (a) We have 
Divide by y
2
Put 
 
? The solution is 
 
7. (c)
? I.F. =  = t = log | x |
solution is given by
y (I.F.) = 
y log | x | =  (log | x |) dx + C = 
8. (c) The equation of normal to a curve at a point (x, y) is
Since it passes throgh the point (3, 0), we have
?