<s>[INST] Initial-value problem $y' = Ax^py$ with initial condition $y(2)=1$ 
Assuming $p\neq-1$, find the solution $y=y(x)$ to the initial-value problem:  
$$\frac{dy}{dx}= Ax^py$$
  $$y(2)=1$$  
in terms of $A$ and $p$.


So far I have, after rearranging the problem and integrating, came up with this equation:
$$\ln(y) = \frac{Ax^{p+1}}{p+1} + C$$
  Simplifying it becomes: 
$$ y = e^{\frac{Ax^{p+1}}{p+1} + C}$$
In which case if you plugged in $y(2) = 1$, $\frac{Ax^{p+1}}{p+1} + C$  must $ = 0$. In which case C must be $=-\frac{Ax^{p+1}}{p+1}$ and the entire solution would be $$y = \frac{Ax^{p+1}}{p+1} -\frac{Ax^{p+1}}{p+1}$$  
However this does not appear to be the correct answer. What am I doing wrong?
 [/INST] You have almost correctly solved the problem except the last step.
$$\ln(y) = \frac{Ax^{p+1}}{p+1} + C$$
Plugging in the value $y(2) = 1$, we get $$\ln(1) = \frac{A\cdot2^{p+1}}{p+1} + C \implies C=-\frac{A\cdot(2)^{p+1}}{p+1}$$
So the required solution is $$\ln(y) = \frac{Ax^{p+1}}{p+1}-\frac{A\cdot2^{p+1}}{p+1}= \frac{A(x^{p+1}-2^{p+1})}{p+1}$$
$$\implies y=\exp\left[\frac{A(x^{p+1}-2^{p+1})}{p+1}\right]$$
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