Question

Solve the given differential equation by separation of variables.

(e^x + e^-x)dy/dx=y^2

Answer 1

Solution to Question #70255, Math / Differential Equations

Question:

Solve the given differential equation by separation of variables.

$$(e ^ {\wedge} x + e ^ {\wedge} - x) d y / d x = y ^ {\wedge} 2$$

Solution:

First let's separate the variables

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

Now for the left part

$$\int \frac {d y}{y ^ {2}} = - \frac {1}{y} + C _ {1}$$

For the right part

$$\int \frac {d x}{(e ^ {x} + e ^ {- x})} = \int \frac {e ^ {x} d x}{((e ^ {x}) ^ {2} + 1)}$$

We will set

$$z = e ^ {x}$$

Then

$$\int \frac {e ^ {x} d x}{((e ^ {x}) ^ {2} + 1)} = \int \frac {d z}{(1 + z ^ {2})} = \arctan z + C _ {2} = \arctan e ^ {x} + C _ {2}$$

Finally

$$- \frac {1}{y} + C _ {1} = \arctan e ^ {x} + C _ {2}$$ $$\frac {1}{y} = - \arctan e ^ {x} - C ^ {*}$$ $$y = - \frac {1}{\arctan e ^ {x} + C ^ {*}}$$

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