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

Finally, taking a logarithm to the simplified equation, we can find the solution $$y=-\ln (-\frac{1}{3}{e}^{3x}+\frac{8}{15}).$$

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