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Question
The differential equation which represents the family of curves y = eCx is
Options
y1 = C2 y
xy1 − ln y = 0
x ln y = yy1
y ln y = xy1
Solution
y ln y = xy1
We have,
y = eCx
Taking ln on both sides, we get
ln y = Cx ln e
⇒ In y = Cx ........(1)
Differentiating both sides of (1) with respect to x, we get
\[\frac{1}{y} y_1 = C\]
Substituting the value of C in (1), we get
\[\ln y = \frac{y_1}{y}x\]
\[ \Rightarrow y \ln y = y_1 x\]
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