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Byju's Answer
Standard XII
Mathematics
Standard Formulae - 3
i d y d x=y t...
Question
(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
Open in App
Solution
(
i
)
d
y
d
x
=
y
tan
x
,
y
0
=
1
⇒
1
y
d
y
=
tan
x
d
x
Integrating
both
sides
,
we
get
∫
1
y
d
y
=
∫
tan
x
d
x
⇒
log
y
=
log
sec
x
+
C
.
.
.
.
.
(
1
)
We
know
that
at
x
=
0
and
y
=
1
.
Substituting
the
values
of
x
and
y
in
(
1
)
,
we
get
log
1
=
log
1
+
C
⇒
C
=
0
Substituting
the
value
of
C
in
(
1
)
,
we
get
log
y
=
log
sec
x
+
0
⇒
y
=
sec
x
Hence
,
y
=
sec
x
,
where
x
∈
-
π
2
,
π
2
,
is
the
required
solution
.
(
ii
)
2
x
d
y
d
x
=
5
y
,
y
1
=
1
⇒
2
y
d
y
=
5
x
d
x
Integrating
both
sides
,
we
get
2
∫
1
y
d
y
=
5
∫
1
x
d
x
⇒
2
log
y
=
5
log
x
+
C
.
.
.
.
.
(
1
)
We
know
that
at
x
=
1
and
y
=
1
.
Substituting
the
values
of
x
and
y
in
(
1
)
,
we
get
2
log
1
=
5
log
1
+
C
⇒
C
=
0
Substituting
the
value
of
C
in
(
1
)
,
we
get
2
log
y
=
5
log
x
+
0
⇒
y
=
x
5
2
Hence
,
y
=
x
5
2
is
the
required
solution
.
(
iii
)
d
y
d
x
=
2
e
2
x
y
2
,
y
0
=
-
1
⇒
1
y
2
d
y
=
2
e
2
x
d
x
Integrating
both
sides
,
we
get
∫
1
y
2
d
y
=
2
∫
e
2
x
d
x
⇒
-
1
y
=
e
2
x
+
C
.
.
.
.
.
(
1
)
We
know
that
at
x
=
0
,
y
=
-
1
.
Substituting
the
values
of
x
and
y
in
(
1
)
,
we
get
1
=
1
+
C
⇒
C
=
0
Substituting
the
value
of
C
in
(
1
)
,
we
get
-
1
y
=
e
2
x
⇒
y
=
-
e
-
2
x
Hence
,
y
=
-
e
-
2
x
is
the
required
solution
.
(
iv
)
cos
y
d
y
d
x
=
e
x
,
y
0
=
π
2
⇒
cos
y
d
y
=
e
x
d
x
Integrating
both
sides
,
we
get
∫
cos
y
d
y
=
∫
e
x
d
x
⇒
sin
y
=
e
x
+
C
.
.
.
.
.
(
1
)
We
know
that
at
x
=
0
,
y
=
π
2
.
Substituting
the
values
of
x
and
y
in
(
1
)
,
we
get
1
=
1
+
C
⇒
C
=
0
Substituting
the
value
of
C
in
(
1
)
,
we
get
sin
y
=
e
x
⇒
y
=
sin
-
1
e
x
Hence
,
y
=
sin
-
1
e
x
is
the
required
solution
.
(
v
)
d
y
d
x
=
2
x
y
,
y
0
=
1
⇒
1
y
d
y
=
2
x
d
x
Integrating
both
sides
,
we
get
∫
1
y
d
y
=
∫
2
x
d
x
log
y
=
x
2
+
C
.
.
.
.
.
(
1
)
We
know
that
at
x
=
0
,
y
=
1
.
Substituting
the
values
of
x
and
y
in
(
1
)
,
we
get
0
=
0
+
C
⇒
C
=
0
Substituting
the
value
of
C
in
(
1
)
,
we
get
log
y
=
x
2
⇒
y
=
e
x
2
Hence
,
y
=
e
x
2
is
the
required
solution
.
(
vi
)
d
y
d
x
=
1
+
x
2
+
y
2
+
x
2
y
2
,
y
0
=
1
⇒
d
y
d
x
=
1
+
x
2
1
+
y
2
⇒
d
y
1
+
y
2
=
1
+
x
2
d
x
Integrating
both
sides
,
we
get
∫
d
y
1
+
y
2
=
∫
1
+
x
2
d
x
⇒
tan
-
1
y
=
x
+
x
3
3
+
C
.
.
.
.
.
(
1
)
We
know
that
at
x
=
0
,
y
=
1
.
Substituting
the
values
of
x
and
y
in
(
1
)
,
we
get
π
4
=
0
+
0
+
C
⇒
C
=
π
4
Substituting
the
value
of
C
in
(
1
)
,
we
get
tan
-
1
y
=
x
+
x
3
3
+
π
4
Hence
,
tan
-
1
y
=
x
+
x
3
3
+
π
4
is
the
required
solution
.
(
vii
)
x
y
d
y
d
x
=
x
+
2
y
+
2
,
y
1
=
-
1
⇒
y
y
+
2
d
y
=
x
+
2
x
d
x
⇒
y
+
2
-
2
y
+
2
d
y
=
x
+
2
x
d
x
⇒
1
-
2
y
+
2
d
y
=
1
+
2
x
d
x
Integrating
both
sides
,
we
get
∫
1
-
2
y
+
2
d
y
=
∫
1
+
2
x
d
x
⇒
y
-
2
log
y
+
2
=
x
+
2
log
x
+
C
.
.
.
.
.
(
1
)
We
know
that
at
x
=
1
,
y
=
-
1
.
Substituting
the
values
of
x
and
y
in
(
1
)
,
we
get
-
1
-
2
log
1
=
1
+
2
log
1
+
C
⇒
-
1
=
1
+
C
⇒
C
=
-
2
Substituting
the
value
of
C
in
(
1
)
,
we
get
y
-
2
log
y
+
2
=
x
+
2
log
x
-
2
Hence
,
y
-
2
log
y
+
2
=
x
+
2
log
x
-
2
is
the
required
solution
.
Suggest Corrections
0
Similar questions
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
(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]
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
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.
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
.
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