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Mathematics
Mathematics
Problem
The solution of the differential equation
d
y
d
x
=
x
+
y
x
\frac{dy}{dx} = \frac{x+y}{x}
d
x
d
y
=
x
x
+
y
satisfying the condition
y
(
1
)
=
1
y(1)=1
y
(
1
)
=
1
is (a)
y
=
log
x
+
x
y = \log x + x
y
=
lo
g
x
+
x
(b)
y
=
x
log
x
+
x
2
y = x \log x + x^2
y
=
x
lo
g
x
+
x
2
(c)
y
=
x
e
(
x
−
1
)
y = xe^{(x-1)}
y
=
x
e
(
x
−
1
)
(d)
y
=
x
log
x
+
x
y = x \log x + x
y
=
x
lo
g
x
+
x
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