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Byju's Answer
Standard XII
Mathematics
Equation of Normal at a Point (x,y) in Terms of f'(x)
Solve each of...
Question
Solve each of the following initial value problems:
(i)
y
'
+
y
=
e
x
,
y
0
=
1
2
(ii)
x
d
y
d
x
-
y
=
log
x
,
y
1
=
0
(iii)
d
y
d
x
+
2
y
=
e
-
2
x
sin
x
,
y
0
=
0
(iv)
x
d
y
d
x
-
y
=
x
+
1
e
-
x
,
y
1
=
0
(v)
1
+
y
2
d
x
+
x
-
e
-
tan
-
1
y
d
x
=
0
,
y
0
=
0
(vi)
d
y
d
x
+
y
tan
x
=
2
x
+
x
2
tan
x
,
y
0
=
1
(vii)
x
d
y
d
x
+
y
=
x
cos
x
+
sin
x
,
y
π
2
=
1
(viii)
d
y
d
x
+
y
cot
x
=
4
x
cosec
x
,
y
π
2
=
0
(ix)
d
y
d
x
+
2
y
tan
x
=
sin
x
;
y
=
0
when
x
=
π
3
(x)
d
y
d
x
-
3
y
cot
x
=
sin
2
x
;
y
=
2
when
x
=
π
2
Open in App
Solution
i
We
have
,
y
'
+
y
=
e
x
⇒
d
y
d
x
+
y
=
e
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
1
and
Q
=
e
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
∫
1
d
x
=
e
x
Multiplying
both
sides
of
1
by
I
.
F
.
=
e
x
,
we
get
e
x
d
y
d
x
+
y
=
e
x
e
x
⇒
e
x
d
y
d
x
+
e
x
y
=
e
2
x
Integrating
both
sides
with
respect
to
x
,
we
get
y
e
x
=
∫
e
2
x
d
x
+
C
⇒
y
e
x
=
e
2
x
2
+
C
.
.
.
.
.
2
Now
,
y
0
=
1
2
∴
1
2
e
0
=
e
0
2
+
C
⇒
C
=
0
Putting
the
value
of
C
in
2
,
we
get
y
e
x
=
e
2
x
2
⇒
e
x
=
e
x
2
Hence
,
y
=
e
x
2
is
the
required
solution
.
ii
We
have
,
x
d
y
d
x
-
y
=
log
x
⇒
d
y
d
x
-
y
x
=
log
x
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
-
1
x
and
Q
=
log
x
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
-
∫
1
x
d
x
=
e
-
log
x
=
1
x
Multiplying
both
sides
of
1
by
I
.
F
.
=
1
x
,
we
get
1
x
d
y
d
x
-
1
x
y
=
1
x
×
log
x
x
⇒
1
x
d
y
d
x
-
1
x
2
y
=
log
x
x
2
Integrating
both
sides
with
respect
to
x
,
we
get
y
1
x
=
∫
1
x
2
II
×
log
x
I
d
x
+
C
⇒
y
x
=
log
x
∫
1
x
2
d
x
-
∫
d
d
x
log
x
∫
1
x
2
d
x
d
x
+
C
⇒
y
x
=
-
log
x
x
+
∫
1
x
2
d
x
+
C
⇒
y
x
=
-
log
x
x
-
1
x
+
C
⇒
y
=
-
log
x
-
1
+
C
x
.
.
.
.
.
2
Now
,
y
1
=
0
∴
0
=
-
0
-
1
+
C
1
⇒
C
=
1
Putting
the
value
of
C
in
2
,
we
get
y
=
-
log
x
-
1
+
x
⇒
y
=
x
-
1
-
log
x
Hence
,
y
=
x
-
1
-
log
x
is
the
required
solution
.
iii
We
have
,
d
y
d
x
+
2
y
=
e
-
2
x
sin
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
2
and
Q
=
e
-
2
x
sin
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
∫
2
d
x
=
e
2
x
Multiplying
both
sides
of
1
by
I
.
F
.
=
e
2
x
,
we
get
e
2
x
d
y
d
x
+
2
y
=
e
2
x
e
-
2
x
sin
x
⇒
e
2
x
d
y
d
x
+
2
y
=
sin
x
Integrating
both
sides
with
respect
to
x
,
we
get
y
e
2
x
=
∫
sin
x
d
x
+
C
⇒
y
e
2
x
=
-
cos
x
+
C
.
.
.
.
.
2
Now
,
y
0
=
0
∴
0
×
e
0
=
-
cos
0
+
C
⇒
C
=
1
Putting
the
value
of
C
in
2
,
we
get
y
e
2
x
=
-
cos
x
+
1
⇒
y
e
2
x
=
1
-
cos
x
Hence
,
y
e
2
x
=
1
-
cos
x
is
the
required
solution
.
iv
We
have
,
x
d
y
d
x
-
y
=
x
+
1
e
-
x
⇒
d
y
d
x
-
1
x
y
=
x
+
1
x
e
-
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
-
1
x
and
Q
=
x
+
1
x
e
-
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
-
∫
1
x
d
x
=
e
-
log
x
=
1
x
Multiplying
both
sides
of
1
by
I
.
F
.
=
1
x
,
we
get
1
x
d
y
d
x
-
1
x
y
=
1
x
x
+
1
x
e
-
x
⇒
1
x
d
y
d
x
-
1
x
y
=
x
+
1
x
2
e
-
x
Integrating
both
sides
with
respect
to
x
,
we
get
1
x
y
=
∫
1
x
+
1
x
2
e
-
x
d
x
+
C
Putting
1
x
e
-
x
=
t
⇒
-
1
x
e
-
x
-
1
x
2
e
-
x
d
x
=
d
t
⇒
1
x
+
1
x
2
e
-
x
d
x
=
-
d
t
∴
1
x
y
=
∫
-
d
t
+
C
⇒
y
x
=
-
t
+
C
⇒
y
x
=
-
e
-
x
x
+
C
⇒
y
=
-
e
-
x
+
C
x
.
.
.
.
.
2
Now
,
y
1
=
0
∴
0
=
-
e
-
1
+
C
⇒
C
=
e
-
1
Putting
the
value
of
C
in
2
,
we
get
y
=
-
e
-
x
+
x
e
-
1
⇒
y
=
x
e
-
1
-
e
-
x
Hence
,
y
=
x
e
-
1
-
e
-
x
is
the
required
solution
.
v
We
have
,
1
+
y
2
d
x
+
x
-
e
-
tan
-
1
y
d
y
=
0
⇒
x
-
e
-
tan
-
1
y
d
y
d
x
=
-
1
+
y
2
⇒
1
+
y
2
d
x
d
y
=
-
x
-
e
-
tan
-
1
y
⇒
d
x
d
y
+
x
1
+
y
2
=
e
-
tan
-
1
y
1
+
y
2
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
x
d
y
+
P
x
=
Q
where
P
=
1
1
+
y
2
and
Q
=
e
-
tan
-
1
y
1
+
y
2
∴
I
.
F
.
=
e
∫
P
d
y
=
e
∫
1
1
+
y
2
d
y
=
e
tan
-
1
y
Multiplying
both
sides
of
1
by
I
.
F
.
=
e
tan
-
1
y
,
we
get
e
tan
-
1
y
d
x
d
y
+
x
1
+
y
2
=
e
tan
-
1
y
e
-
tan
-
1
y
1
+
y
2
⇒
e
tan
-
1
y
d
x
d
y
+
x
1
+
y
2
=
1
1
+
y
2
Integrating
both
sides
with
respect
to
y
,
we
get
e
tan
-
1
y
x
=
∫
1
1
+
y
2
d
y
+
C
⇒
x
e
tan
-
1
y
=
tan
-
1
y
+
C
.
.
.
.
.
2
Now
,
y
0
=
0
∴
0
×
e
0
=
0
+
C
⇒
C
=
0
Putting
the
value
of
C
in
2
,
we
get
x
e
tan
-
1
y
=
tan
-
1
y
+
0
⇒
x
e
tan
-
1
y
=
tan
-
1
y
Hence
,
x
e
tan
-
1
y
=
tan
-
1
y
is
the
required
solution
.
vi
We
have
,
d
y
d
x
+
y
tan
x
=
2
x
+
x
2
tan
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
tan
x
and
Q
=
x
2
cot
x
+
2
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
∫
tan
x
d
x
=
e
log
sec
x
=
sec
x
Multiplying
both
sides
of
1
by
I
.
F
.
=
sec
x
,
we
get
sec
x
d
y
d
x
+
y
tan
x
=
sec
x
x
2
tan
x
+
2
x
⇒
sec
x
d
y
d
x
+
y
tan
x
=
x
2
tan
x
sec
x
+
2
x
sec
x
Integrating
both
sides
with
respect
to
x
,
we
get
y
sec
x
=
∫
x
2
tan
x
sec
x
d
x
+
2
∫
x
II
sec
x
I
d
x
+
C
⇒
y
sec
x
=
∫
x
2
tan
x
sec
x
d
x
+
2
s
e
c
x
∫
x
d
x
-
2
∫
d
d
x
s
e
c
x
∫
x
d
x
d
x
+
C
⇒
y
sec
x
=
∫
x
2
tan
x
sec
x
d
x
+
x
2
sec
x
-
∫
x
2
tan
x
sec
x
d
x
+
C
⇒
y
sec
x
=
x
2
sec
x
+
C
⇒
y
=
x
2
+
C
cos
x
.
.
.
.
.
2
Now
,
y
0
=
1
∴
1
=
0
+
C
cos
0
⇒
C
=
1
Putting
the
value
of
C
in
2
,
we
get
y
=
x
2
+
cos
x
Hence
,
y
=
x
2
+
cos
x
is
the
required
solution
.
vii
We
have
,
x
d
y
d
x
+
y
=
x
cos
x
+
sin
x
⇒
d
y
d
x
+
1
x
y
=
cos
x
+
sin
x
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
tan
x
and
Q
=
x
2
cot
x
+
2
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
∫
1
x
d
x
=
e
log
x
=
x
Multiplying
both
sides
of
(
1
)
by
I
.
F
.
=
x
,
we
get
x
d
y
d
x
+
1
x
y
=
x
cos
x
+
sin
x
x
⇒
x
d
y
d
x
+
1
x
y
=
x
cos
x
+
sin
x
Integrating
both
sides
with
respect
to
x
,
we
get
x
y
=
∫
x
cos
x
d
x
+
∫
sin
x
d
x
+
C
⇒
x
y
=
x
sin
x
-
∫
1
sin
x
d
x
-
cos
x
+
C
⇒
x
y
=
x
sin
x
+
cos
x
-
cos
x
+
C
⇒
x
y
=
x
sin
x
+
C
.
.
.
.
.
2
Now
,
y
π
2
=
1
∴
1
×
π
2
=
π
2
sin
π
2
+
C
⇒
C
=
0
Putting
the
value
of
C
in
2
,
we
get
x
y
=
x
sin
x
⇒
y
=
sin
x
Hence
,
y
=
sin
x
is
the
required
solution
.
viii
We
have
,
d
y
d
x
+
y
cot
x
=
4
x
cosec
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
cot
x
and
Q
=
4
x
cosec
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
∫
cot
x
d
x
=
e
log
sin
x
=
sin
x
Multiplying
both
sides
of
1
by
I
.
F
.
=
sin
x
,
we
get
sin
x
d
y
d
x
+
y
cot
x
=
sin
x
4
x
cosec
x
⇒
sin
x
d
y
d
x
+
y
cot
x
=
4
x
Integrating
both
sides
with
respect
to
x
,
we
get
y
sin
x
=
4
∫
x
d
x
+
C
⇒
y
sin
x
=
2
x
2
+
C
.
.
.
.
.
2
Now
,
y
π
2
=
0
∴
0
×
sin
π
2
=
2
π
2
2
+
C
⇒
C
=
-
π
2
2
Putting
the
value
of
C
in
2
,
we
get
y
sin
x
=
2
x
2
-
π
2
2
Hence
,
y
sin
x
=
2
x
2
-
π
2
2
is
the
required
solution
.
ix
We
have
,
d
y
d
x
+
2
y
tan
x
=
sin
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
2
tan
x
and
Q
=
sin
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
2
∫
tan
x
d
x
=
e
2
log
sec
x
=
sec
2
x
Multiplying
both
sides
of
(
1
)
by
I
.
F
.
=
sec
2
x
,
we
get
sec
2
x
d
y
d
x
+
2
y
tan
x
=
sec
2
x
×
sin
x
⇒
sec
2
x
d
y
d
x
+
2
y
tan
x
=
tan
x
sec
x
Integrating
both
sides
with
respect
to
x
,
we
get
y
sec
2
x
=
∫
tan
x
sec
x
d
x
+
C
⇒
y
sec
2
x
=
sec
x
+
C
.
.
.
.
.
2
Now
,
y
π
3
=
0
∴
0
sec
π
3
2
=
sec
π
3
+
C
⇒
C
=
-
2
Putting
the
value
of
C
in
2
,
we
get
y
sec
2
x
=
sec
x
-
2
⇒
y
=
cos
x
-
2
cos
2
x
Hence
,
y
=
cos
x
-
2
cos
2
x
is
the
required
solution
.
x
We
have
,
d
y
d
x
-
3
y
cot
x
=
sin
2
x
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
y
d
x
+
P
y
=
Q
where
P
=
-
3
cot
x
and
Q
=
sin
2
x
∴
I
.
F
.
=
e
∫
P
d
x
=
e
-
3
∫
cot
x
d
x
=
e
-
3
log
sin
x
=
cosec
3
x
Multiplying
both
sides
of
1
by
I
.
F
.
=
cosec
3
x
,
we
get
cosec
3
x
d
y
d
x
-
3
y
cot
x
=
sin
2
x
cosec
3
x
⇒
cosec
3
x
d
y
d
x
-
3
y
cot
x
=
2
cot
x
cosec
x
Integrating
both
sides
with
respect
to
x
,
we
get
y
cosec
3
x
=
2
∫
cot
x
cosec
x
d
x
+
C
⇒
y
cosec
3
x
=
-
2
cosec
x
+
C
⇒
y
=
-
2
sin
2
x
+
C
sin
3
x
.
.
.
.
.
2
Now
,
y
π
2
=
2
∴
2
=
-
2
sin
2
π
2
+
C
sin
3
π
2
⇒
C
=
4
Putting
the
value
of
C
in
2
,
we
get
y
=
-
2
sin
2
x
+
4
sin
3
x
⇒
y
=
4
sin
3
x
-
2
sin
2
x
Hence
,
y
=
4
sin
3
x
-
2
sin
2
x
is
the
required
solution
.
Suggest Corrections
0
Similar questions
Q.
Solve each of the following initial value problems:
(i)
y
'
+
y
=
e
x
,
y
0
=
1
2
(ii)
x
d
y
d
x
-
y
=
log
x
,
y
1
=
0
(iii)
d
y
d
x
+
2
y
=
e
-
2
x
sin
x
,
y
0
=
0
(iv)
x
d
y
d
x
-
y
=
x
+
1
e
-
x
,
y
1
=
0
(v)
1
+
y
2
d
x
+
x
-
e
-
tan
-
1
y
d
x
=
0
,
y
0
=
0
(vi)
d
y
d
x
+
y
tan
x
=
2
x
+
x
2
tan
x
,
y
0
=
1
(vii)
x
d
y
d
x
+
y
=
x
cos
x
+
sin
x
,
y
π
2
=
1
(viii)
d
y
d
x
+
y
cot
x
=
4
x
cosec
x
,
y
π
2
=
0
(ix)
d
y
d
x
+
2
y
tan
x
=
sin
x
;
y
=
0
when
x
=
π
3
(x)
d
y
d
x
-
3
y
cot
x
=
sin
2
x
;
y
=
2
when
x
=
π
2
(xi)
d
y
d
x
+
y
cot
x
=
2
cos
x
,
y
π
2
=
0
(xii)
d
y
=
cos
x
2
-
y
cos
e
c
x
d
x
(xiii)
tan
x
d
y
d
x
=
2
x
tan
x
+
x
2
-
y
;
tan
x
≠
0
given that y = 0 when
x
=
π
2
.
Q.
Solve each of the following initial value problems:
(i)
y
'
+
y
=
e
x
,
y
0
=
1
2
(ii)
x
d
y
d
x
-
y
=
log
x
,
y
1
=
0
(iii)
d
y
d
x
+
2
y
=
e
-
2
x
sin
x
,
y
0
=
0
(iv)
x
d
y
d
x
-
y
=
x
+
1
e
-
x
,
y
1
=
0
(v)
1
+
y
2
d
x
+
x
-
e
-
tan
-
1
y
d
x
=
0
,
y
0
=
0
(vi)
d
y
d
x
+
y
tan
x
=
2
x
+
x
2
tan
x
,
y
0
=
1
(vii)
x
d
y
d
x
+
y
=
x
cos
x
+
sin
x
,
y
π
2
=
1
(viii)
d
y
d
x
+
y
cot
x
=
4
x
cosec
x
,
y
π
2
=
0
(ix)
d
y
d
x
+
2
y
tan
x
=
sin
x
;
y
=
0
when
x
=
π
3
(x)
d
y
d
x
-
3
y
cot
x
=
sin
2
x
;
y
=
2
when
x
=
π
2
(xi)
(xii)
Q.
(i)
d
y
d
x
=
y
tan
x
,
y
0
=
1
(ii)
2
x
d
y
d
x
=
5
y
,
y
1
=
1
(iii)
d
y
d
x
=
2
e
2
x
y
2
,
y
0
=
-
1
(iv)
cos
y
d
y
d
x
=
e
x
,
y
0
=
π
2
(v)
d
y
d
x
=
2
x
y
,
y
0
=
1
(vi)
d
y
d
x
=
1
+
x
2
+
y
2
+
x
2
y
2
,
y
0
=
1
(vii)
x
y
d
y
d
x
=
x
+
2
y
+
2
,
y
1
=
-
1
(viii)
d
y
d
x
=
1
+
x
+
y
2
+
x
y
2
when y = 0, x = 0 [NCERT EXEMPLAR]
(ix)
2
y
+
3
-
x
y
d
y
d
x
=
0
, y(1) = −2 [NCERT EXEMPLAR]
Q.
(i)
d
y
d
x
=
y
tan
x
,
y
0
=
1
(ii)
2
x
d
y
d
x
=
5
y
,
y
1
=
1
(iii)
d
y
d
x
=
2
e
2
x
y
2
,
y
0
=
-
1
(iv)
cos
y
d
y
d
x
=
e
x
,
y
0
=
π
2
(v)
d
y
d
x
=
2
x
y
,
y
0
=
1
(vi)
d
y
d
x
=
1
+
x
2
+
y
2
+
x
2
y
2
,
y
0
=
1
(vii)
x
y
d
y
d
x
=
x
+
2
y
+
2
,
y
1
=
-
1
Q.
(i)
d
y
d
x
=
y
tan
x
,
y
0
=
1
(ii)
2
x
d
y
d
x
=
5
y
,
y
1
=
1
(iii)
d
y
d
x
=
2
e
2
x
y
2
,
y
0
=
-
1
(iv)
cos
y
d
y
d
x
=
e
x
,
y
0
=
π
2
(v)
d
y
d
x
=
2
x
y
,
y
0
=
1
(vi)
d
y
d
x
=
1
+
x
2
+
y
2
+
x
2
y
2
,
y
0
=
1
(vii)
x
y
d
y
d
x
=
x
+
2
y
+
2
,
y
1
=
-
1
(viii)
d
y
d
x
=
1
+
x
+
y
2
+
x
y
2
when y = 0, x = 0 [NCERT EXEMPLAR]
(ix)
2
y
+
3
-
x
y
d
y
d
x
=
0
, y(1) = −2 [NCERT EXEMPLAR]
(x)
e
x
tan
y
d
x
+
2
-
e
x
sec
2
y
d
y
=
0
,
y
0
=
π
4
[CBSE 2018]
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