# Solution to h'(t) = h(t)^j : Wolfram Alpha mistake?

This should be a quick and easy one. Trying to teach myself a bit more about differential equations, so I put the following equation into Wolfram Alpha:

$$h'(t) = h(t)^j$$

It gave me the following output:

$$h(t) = [(j-1)(c-t)]^{\frac{1}{1-j}}$$

After checking, it seems like the correct answer should be:

$$h(t) = [(j-1)(c+t)]^{\frac{1}{1-j}}$$

So, did Wolfram Alpha goof here? Or is there something I am not understanding?

The equation is separable $$\frac{dt}{dh}=h^{-j}$$ So $$t+c_1=\frac{h^{1-j}}{1-j}$$ Solving for $h$ gives in the most general case $$h=\Big((j-1) (-c_1-t)\Big)^{\frac{1}{1-j}}=\Big((j-1) (c_2-t)\Big)^{\frac{1}{1-j}}$$ which is what Wolfram Alpha returns.