Question

solve the folowing differential equation y'cosx= y/lny

Answer 1

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.