Question
solve the folowing differential equation y'cosx= y/lny
Answers
GPT-4o mini
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:
1. Separate the variables:
\[
\frac{dy}{y} = \sec x \, dx.
\]
2. 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 \(C = C_2 - C_1\).
3. 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.
\[
y' \cos x = \frac{y}{\ln y},
\]
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:
1. Separate the variables:
\[
\frac{dy}{y} = \sec x \, dx.
\]
2. 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 \(C = C_2 - C_1\).
3. 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.