<s>[INST] Finding constant $a$ so that differential equation becomes correct Question:
Find the value of $a$ so that the function
$$y = \sqrt{x} \ln{x}$$
is a solution to the differential equation
$$y' - \frac{a}{x} \cdot y = \frac{1}{\sqrt{x}}$$
Attempted solution:
My basic approach would be to take the derivative of y, put y and y' into the differential equation and solve for a.
The derivative of y is gotten with the product rule and then the chain rule for the square root:
$$y' = \sqrt{x} \cdot (\ln{x})' + (\sqrt{x})' \cdot \ln{x} = \frac{\sqrt{x}}{x} + \frac{\ln{x}}{2 \cdot \sqrt{x}}$$
Solving for a:
$$y' - \frac{a}{x} \cdot y = \frac{1}{\sqrt{x}} \Rightarrow a = \frac{y' - \frac{1}{\sqrt{x}}}{y} = \frac{xy' - \sqrt{x}}{y}$$
Putting y and y' into the expression for a gives:
$$a = \frac{x(\frac{\sqrt{x}}{x} + \frac{\ln {x}}{2\sqrt{x}})}{\sqrt{x}\ln {x}} = \frac{1+\frac{\ln{x}}{2}}{\ln{x}} = \frac{1}{\ln{x}} + \frac{1}{2}$$
The expected answer is just $\frac{1}{2}$, so for some reason I have gotten an extra $\frac{1}{\ln{x}}$ somwhere.
 [/INST] It is $$\frac{\ln(x)}{2\sqrt{x}}+\frac{\sqrt{x}}{x}-\frac{a}{x}\sqrt{x}\ln(x)=\frac{1}{\sqrt{x}}$$ multiplying by $\sqrt{x}$ we get
$$\frac{\ln(x)}{2}-a\ln(x)=0$$ so $$a=\frac{1}{2}$$ for $\ln(x)\ne 0$
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