Question

Solve the equation\({e^{ - y}}{y^\prime } + \cos x = 0\)and graph several members of the family of solutions. How does the solution curve change as the constant\(C\)varies?

Short answer

\(y = \ln \frac{1}{{(\sin + C)}}\), Yes

Step-by-step solution

Definition

Separable differential equations can be written in the form\(\frac{{dy}}{{dx}} = {\rm{ }}f\left( x \right){\rm{ }}g\left( y \right)\), where\(x,y\)are the variables and are explicitly separated from each other.

Separate variables

We have to solve\({e^{ - y}}{y^\prime } + \cos x = 0\)

\(\begin{aligned}{}{e^{ - y}}\frac{{dy}}{{dx}} + \cos x = 0\\{e^{ - y}}dy = - \cos xdx\end{aligned}\)

Integrate

Integrating both sides:

\(\begin{aligned}\int {{e^{ - y}}} dy &= - \int {\cos } xdx\\ - {e^{ - y}} &= - \sin x + {C_1}\\{e^{ - y}} &= \sin x + C\left( {C = - {C_1}} \right)\\ - y &= \ln (\sin x + C)\\y &= \ln \frac{1}{{(\sin + C)}}\end{aligned}\)

Graph

The graph using graphing calculator is:

The Curve is not defined when.

For, curve is continuous.

We see that as increases, graph of solution shifts downward and amplitude of the oscillations decreases.