Finding one function that can satisfy both of these equations I have two equations as follows:
$$2f'(x)-\sqrt x\ln x=0$$
$$f(e)=0$$
But I'm not sure how I can find one function $f$ to satisfy both since the second equation has different variables. I've tried isolating $f'(x)$ on the first one, and then trying to find $f$ through integration but I don't think that's right.
Thanks  :)
 A: You need to integrate the first equation (indefinite integration) and remeber to include constant in your answer.
Then put the condition of second equation to find the value of constant.
F(x) = ((ln(x) * x^(3/2)) - (2 * x^(1/2))) /2+ c
Putting x = e, satisfy f(x) = 0.
Thus, c = - e^(1/2) * (e - 2) * (1/2)
Note: beware of the brackets.
A: Integration is the right way to go for this problem.
$$2f'(x)-\sqrt x\ln x=0$$
$$f'(x)=\frac12\sqrt x\ln x$$
$$f(x)=\int\frac12\sqrt x\ln x\,dx$$
$$=\frac12\left(\frac23x^{3/2}\ln x
-\int\frac{\frac23x^{3/2}}x\,dx\right)$$
$$=\frac12\left(\frac23x^{3/2}\ln x-\frac49x^{3/2}+K\right)$$
$$=\frac13x^{3/2}\ln x-\frac29x^{3/2}+K$$
Since $f(e)=0$:
$$0=\frac13e^{3/2}\ln e-\frac29e^{3/2}+K$$
$$K=\frac29e^{3/2}-\frac13e^{3/2}=-\frac19e^{3/2}$$
Finally we have
$$f(x)=\frac13x^{3/2}\ln x-\frac29x^{3/2}-\frac19e^{3/2}$$
A: $$f(x) =\frac{1}{2}\int \sqrt{x}\ln xdx=\frac{1}{3}x^\frac{3}{2}\ln x-\frac{1}{3}\int x^\frac{1}{2}dx=\frac{1}{3}x^\frac{3}{2}\ln x-\frac{2}{9}x^\frac{3}{2}+C$$
Using the constraint $f(e) =0$, $C=\frac{-2}{9}e^\frac{3}{2}$.
