Question

Solve the given differential equation by separation of variables.

$\left(e^x+e^{-x}\right)\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}{\mathit{y}}^{\mathbf{2}}$

Short answer

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

Step-by-step solution

Definition of separable equation.

A first-order differential equation of the form

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{h}\mathbf{\left(}\mathit{y}\mathbf{\right)}$

is said to be separable or to have separable variables.

Separate the variables.

$\left(e^x+e^{-x}\right)\frac{dy}{dx}=y^2$

Separate the variables

$\frac{dy}{y^2}=\frac{dx}{e^x+e^{-x}}$

Rewrite

role="math" $e^{-x}=\frac{1}{e^{x^{\text{'}}}}$

So

$\frac{dy}{y^2}=\frac{dx}{e^x+\frac{1}{e^x}}$

Simplify

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

Step 3: Integration.

Integrate both sides

$\int \frac{dy}{y^2}=\int \frac{e^{x}dx}{{\left(e^x\right)}^2}+1$

Remember that

$$\begin{gathered} \int \frac{du}{u^2+1}=\arctan u+C \\ So \\ -\frac{1}{y}=\arctan e^x+C \end{gathered}$$

Solve for

So

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