MCQ
The solution of $\frac{{dy}}{{dx}} + \frac{y}{3} = 1$ is
- A$y = 3 + c{e^{x/3}}$
- ✓$y = 3 + c{e^{ - x/3}}$
- C$3y = c + {e^{x/3}}$
- D$3y = c + {e^{ - x/3}}$
✓
Answer
Correct option: B.
$y = 3 + c{e^{ - x/3}}$
b
(b) Given, $\frac{{dy}}{{dx}} + \frac{y}{3} = 1$; $I.F.$ = ${e^{\int {\frac{1}{3}\,d\,x} }} = {e^{x/3}}$
(b) Given, $\frac{{dy}}{{dx}} + \frac{y}{3} = 1$; $I.F.$ = ${e^{\int {\frac{1}{3}\,d\,x} }} = {e^{x/3}}$
Hence, solution is $y\,.\,{e^{x/3}} = \int {1\,.\,{e^{x/3}}\,dx + c} $
$y\,.\,{e^{x/3}} = 3\,{e^{x/3}} + c$; $y = 3 + c{e^{ - x/3}}$.
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