Question

In Problems 1–22 solve the given differential equation by separation

of variables.

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}}$

Short answer

The solution of the given differential equation is$y=-\frac{\ln \left(-\frac{2}{3}{e}^{3x}+c\right)}{2}$

Step-by-step solution

Step 1:Separable Equation.

A first-order differential equation of the form

$\frac{dy}{dx}=g\left(x\right)h\left(y\right)$

is said to be separable or to have separable variables.

Separate the variables and integrate.

The given equation is, $\frac{dy}{dx}=e^{3x+2y}$.

Separate the variables, $e^{-2y}dy=e^{3x}dx\left(\frac{1}{e^n}=e^{-n}\right)$.

Integrate both sides,

$$\begin{gathered} \int e^{-2y}dy=\int e^{3x}dx \\ -\frac{1}{2}{e}^{-2y}=\frac{1}{3}{e}^{3x}+c \\ {e}^{-2y}=-\frac{2}{3}{e}^{3x}+c \end{gathered}$$

Solve for y , so

$\begin{array}{rcl}-2y& =& \ln \left(-\frac{2}{3}{e}^{3x}+c\right)\\ y& =& -\frac{\ln \left(-\frac{2}{3}{e}^{3x}+c\right)}{2}\end{array}$

Hence, the solution of the given differential equation is $y=-\frac{\ln \left(-\frac{2}{3}{e}^{3x}+c\right)}{2}$.