To solve the differential equation
\[ y' \cos x = \frac{y}{\ln y}, \]
we can rewrite it in a more standard form. Starting from the given equation, we can express \(y'\) (which is \(\frac{dy}{dx}\)):
\[ \frac{dy}{dx} = \frac{y}{\ln y} \cdot \frac{1}{\cos x}. \]
Now, we can separate variables. Reorganizing gives us:
\[ \ln y , dy = y \sec x , dx, \]
where \(\sec x = \frac{1}{\cos x}\).
Next, we can proceed by integrating both sides. However, this requires us to manipulate the left side for integration:
- Separate the variables:
\[ \frac{dy}{y} = \sec x , dx. \]
- Now integrate both sides:
Integrating the left side gives:
\[ \int \frac{dy}{y} = \ln |y| + C_1. \]
Integrating the right side requires the integral of \(\sec x\):
\[ \int \sec x , dx = \ln | \sec x + \tan x | + C_2. \]
Putting it together, we have:
\[ \ln |y| = \ln | \sec x + \tan x | + C, \]
where \(C = C_2 - C_1\).
- Exponentiating both sides to solve for \(y\):
\[ |y| = e^C | \sec x + \tan x |. \]
Let \(k = e^C\), which is a constant:
\[ y = k |\sec x + \tan x|. \]
Thus, the general solution to the differential equation is given by:
\[ y = k (\sec x + \tan x), \]
where \(k\) is a constant. The absolute value can often be omitted if we consider both positive and negative \(k\).
Conclusion
The general solution to the differential equation \(y' \cos x = \frac{y}{\ln y}\) is
\[ y = k (\sec x + \tan x), \]
where \(k\) is a constant determined by initial conditions, if provided.
3 answers