Differential Equations Exercise 28
We want to solve the following first-order differential equation:
$$ \begin{cases} y'=\cos^2 y \\ \\ y(0)=0 \end{cases} $$
This is a Cauchy problem since we’re given specific initial conditions.
To find the general solution, I’ll use the separation of variables method, because the equation has the form y=a(x)b(y) with a(x)=1 and b(y)=cos2(y).
$$ y' = \cos^2 y $$
First, rewrite y' in differential form:
$$ \frac{dy}{dx} = \cos^2 y $$
Now separate the variables:
$$ \frac{1}{\cos^2 y} \ dy = \ dx $$
Integrating both sides gives:
$$ \int \frac{1}{\cos^2 y} \ dy = \int \ dx $$
The right-hand side evaluates to x + c:
$$ \int \frac{1}{\cos^2 y} \ dy = x+c $$
The left-hand side is simply tan(y):
$$ \tan(y) = x+c $$
Applying the initial condition x=0, y=0 to solve for c:
$$ \tan(0) = 0+c $$
Since tan(0)=0, we obtain:
$$ c = 0 $$
Therefore, with c=0, the solution to the differential equation is:
$$ \tan(y) = x $$
Taking the arctangent of both sides yields:
$$ y = \arctan(x) $$
This is the solution to the Cauchy problem.
Note. Let’s check whether the equation admits constant solutions. Suppose x=0 and y=0: $$ y'=\cos^2 y $$ $$ y'=\cos^2 0 $$ $$ y'=1 $$ Since the derivative is not zero, the equation does not allow constant solutions.
And that completes the problem.
