Science:Math Exam Resources/Courses/MATH105/April 2018/Question 02 (c)/Solution 1

Since the given equation is separable, we can rewrite is as

$${\frac {dy}{e^{y}}}=e^{3x}dx.$$

Taking integral on the both side of the equation, we have

$$\int e^{-y}dy=\int e^{3x}dx.$$

Note that using substitution $u=ax$ for any fixed number $a\neq 0$, we have

$$\int e^{ax}dx=\int e^{u}{\frac {du}{a}}={\frac {1}{a}}e^{u}+C={\frac {1}{a}}e^{ax}+C.$$

Applying this for $a=-1$ and $a=3$, we obtain

$$\int e^{-y}dy={\frac {1}{-1}}e^{-y}+C_{1}=-e^{-y}+C_{1},$$

and

$$\int e^{3x}dx={\frac {1}{3}}e^{3x}+C_{2},$$

where $C_{1}$ and $C_{2}$ are arbitrary constants.

Therefore, we get

$$-e^{-y}+C_{1}=\int e^{-y}dy=\int e^{3x}dx={\frac {1}{3}}e^{3x}+C_{2},$$

which can be simplified as follows

$$e^{-y}=-{\frac {1}{3}}e^{3x}+C.$$

We plug $x=0$ to find the constant $C$,

$${\frac {1}{5}}=e^{-y(0)}=-{\frac {1}{3}}+C\iff C={\frac {8}{15}}.$$

Finally, taking a logarithm to the simplified equation, we can find the solution

$$y=-\ln \left(-{\frac {1}{3}}e^{3x}+{\frac {8}{15}}\right).$$

Answer: $\color {blue}y=-\ln \left(-{\frac {1}{3}}e^{3x}+{\frac {8}{15}}\right).$