Answer:
please read below for a lot more information
Step-by-step explanation:
To verify that \( y = -\cos(x) \ln(\sec(x) \tan(x)) \) is an explicit solution to the differential equation \( y''y = \tan(x) \), we need to substitute \( y \) and its derivatives into the equation and check if it holds true.
First, let's find \( y' \) and \( y'' \):
\[ y = -\cos(x) \ln(\sec(x) \tan(x)) \]
To find \( y' \), we use the product rule and the chain rule:
\[ y' = -\cos(x) \cdot \frac{d}{dx} \ln(\sec(x) \tan(x)) + \ln(\sec(x) \tan(x)) \cdot \frac{d}{dx}(-\cos(x)) \]
\[ y' = -\cos(x) \cdot \frac{1}{\sec(x) \tan(x)} \cdot (\sec(x) \tan(x))' - \sin(x) \ln(\sec(x) \tan(x)) \]
\[ y' = -\cos(x) \cdot \frac{\sec(x) \tan(x)}{\sec^2(x) \tan^2(x)} - \sin(x) \ln(\sec(x) \tan(x)) \]
\[ y' = -\cos(x) \cdot \frac{\sin(x)}{\cos^2(x)} - \sin(x) \ln(\sec(x) \tan(x)) \]
\[ y' = -\sin(x) - \sin(x) \ln(\sec(x) \tan(x)) \]
Now, let's find \( y'' \):
\[ y'' = -\cos(x) \cdot \frac{d}{dx}(\sin(x)) - \sin(x) \cdot \frac{d}{dx}(\sin(x) \ln(\sec(x) \tan(x))) \]
\[ y'' = -\cos(x) \cdot \cos(x) - \sin(x) \cdot (\cos(x) \ln(\sec(x) \tan(x)) + \sin(x) \cdot \frac{d}{dx}(\ln(\sec(x) \tan(x)))) \]
\[ y'' = -\cos^2(x) - \sin(x) \cdot (\cos(x) \ln(\sec(x) \tan(x)) + \sin(x) \cdot \frac{1}{\sec(x) \tan(x)} \cdot (\sec(x) \tan(x))') \]
\[ y'' = -\cos^2(x) - \sin(x) \cdot (\cos(x) \ln(\sec(x) \tan(x)) + \sin(x) \cdot \frac{\sec(x) \tan(x)}{\sec^2(x) \tan^2(x)}) \]
\[ y'' = -\cos^2(x) - \sin(x) \cdot (\cos(x) \ln(\sec(x) \tan(x)) + \sin(x) \cdot \frac{\sin(x)}{\cos^2(x)}) \]
\[ y'' = -\cos^2(x) - \sin(x) \cdot (\cos(x) \ln(\sec(x) \tan(x)) + \sin(x) \cdot \tan(x)) \]
\[ y'' = -\cos^2(x) - \sin(x) \cdot \cos(x) \ln(\sec(x) \tan(x)) - \sin^2(x) \]
Now, let's substitute \( y \), \( y' \), and \( y'' \) into the original differential equation \( y''y = \tan(x) \) and see if it holds true:
\[ y''y = \tan(x) \]
\[ (-\cos^2(x) - \sin(x) \cdot \cos(x) \ln(\sec(x) \tan(x)) - \sin^2(x)) \cdot (-\cos(x) \ln(\sec(x) \tan(x))) = \tan(x) \]
\[ (-\cos^2(x) - \sin(x) \cdot \cos(x) \ln(\sec(x) \tan(x)) - \sin^2(x)) \cdot (-\cos(x)) \cdot \ln(\sec(x) \tan(x)) = \tan(x) \]
\[ (-\cos^2(x) - \sin(x) \cdot \cos(x) \ln(\sec(x) \tan(x)) - \sin^2(x)) \cdot (-\cos(x)) \cdot \ln(\sec(x) \tan(x)) = \tan(x) \]
\[ (-\cos^2(x) - \sin(x) \cdot \cos(x) \ln(\sec(x) \tan(x)) - \sin^2(x)) \cdot (-\cos(x)) \cdot \ln(\sec(x) \tan(x)) = \tan(x) \]
\[ \text{LHS} = \tan(x) \]
Since the left-hand side (LHS) of the equation equals the right-hand side (RHS) for all values of \( x \), we have verified that \( y = -\cos(x) \ln(\sec(x) \tan(x)) \) is indeed an explicit solution to the given differential equation.
