Engineering
Mathematics
Variable Separable Differential Equation
Question
If \(\frac{{dy}}{{dx}} = y + 3 > 0\) and y (0) = 2, then y (ln 2) is equal to :
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Solution
\(\frac{{dy}}{{dx}} = y + 3\)
\(\frac{{dy}}{{y + 3}} = dx\)
ln(y + 3) = x + c
given at x = 0, y = 2
ln5 = c
\ ln(y + 3) = x + ln5
\(\ln \left( {\frac{{y + 3}}{5}} \right) = x\)
y + 3 = 5ex
y + 5ex – 3
\ y(ln2) = 5eln2 – 3 = 7