y^prime=((1)/(x)+x^x) y^x

To solve the differential equation \( y' = \left(\frac{1}{x} + x^x\right) y^x \), we can use the method of separation of variables. Let's separate the variables \( y \) and \( x \) on different sides of the equation.<br /><br />First, rewrite the equation as:<br />\[ y' = \left(\frac{1}{x} + x^x\right) y^x \]<br /><br />Separate the variables by dividing both sides by \( y^x \):<br />\[ \frac{1}{y^x} \, dy = \left(\frac{1}{x} + x^x\right) \, dx \]<br /><br />Now, integrate both sides:<br />\[ \int \frac{1}{y^x} \, dy = \int \left(\frac{1}{x} + x^x\right) \, dx \]<br /><br />The left side can be simplified using the power rule for integration:<br />\[ \int y^{-x} \, dy = \int y^{-x} \, dy = \frac{y^{-x}}{-x} = -\frac{1}{x} y^{-x} \]<br /><br />For the right side, we need to integrate term by term:<br />\[ \int \frac{1}{x} \, dx + \int x^x \, dx \]<br /><br />The first term is straightforward:<br />\[ \int \frac{1}{x} \, dx = \ln|x| \]<br /><br />The second term involves an exponential function with a variable base, which can be solved using the substitution method. Let \( u = x^x \), then \( \ln u = x \ln x + \ln x \), and \( du = (x \ln x + \ln x) \, dx \). However, this term is more complex and typically requires advanced techniques or numerical methods to solve.<br /><br />Combining the results, we get:<br />\[ -\frac{1}{x} y^{-x} = \ln|x| + \int x^x \, dx + C \]<br /><br />where \( C \) is the constant of integration.<br /><br />This is the general solution to the given differential equation.