Question

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

of variables.

$\mathit{d}\mathit{x}\mathbf{+}{\mathit{e}}^{\mathbf{3}\mathbf{x}}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution of the given differential equation is.$y=\frac{1}{3}{e}^{-3x}+C$

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,$dx+e^{3x}dy=0$.

Rearrange,

$e^{3x}dy=-dx$

Separate the variables,$dy=-\frac{dx}{e^{3x}}$.

Apply$\frac{1}{e^n}=e^{-n}$,so

$dy=-e^{-3x}dx$

Integrate both sides,

$\int dy=\int (-e^{-3x})dx$

Multiply and divide by 3 the second integral,

$\int dy=\frac{1}{3}\int e^{-3x}(-3)dx$

Remember that, $\int e^{u}du=e^u+C$,so

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

Hence, the solution of the given differential equation is.role="math" $\mathit{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}\mathbf{+}\mathit{C}$