Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The general solution of the differential equation x y' +y=x2, x > 0 is
Q. The general solution of the differential equation
x
y
′
+
y
=
x
2
,
x
>
0
is
2163
176
KEAM
KEAM 2020
Report Error
A
y
=
2
x
2
+
C
x
B
y
=
3
x
3
+
C
C
y
=
3
x
2
+
C
D
y
=
3
x
3
+
x
C
E
y
=
3
x
2
+
x
C
Solution:
d
x
d
y
+
x
y
=
x
I
⋅
F
=
e
∫
x
1
d
x
=
e
l
o
g
x
=
x
y
x
=
∫
x
⋅
x
⋅
d
x
y
x
=
3
x
3
+
C
y
=
3
x
2
+
x
c