<s>[INST] Find expression in terms of x Knowing that $$\frac{dy}{dx}= k\cdot x^{\frac{1}{3}}$$ and given that it passes through points $(1,4)$ and $(8,16)$, find an expression for the path in terms of $x$. 
I found out that $$y= \frac34 k x^{\frac43}$$ by integrating $$\frac{dy}{dx}.$$ 
What do I do next? 
 [/INST] We have, $$\frac{dy}{dx}=kx^{\frac{1}{3}}$$ $$\implies dy=kx^{\frac{1}{3}}dx$$ Integrating both the sides w.r.t. $x$ as follows $$\int dy=\int kx^{\frac{1}{3}}dx$$ $$\implies y=k\int x^{\frac{1}{3}}dx=k\left(\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right)+C$$ $$\implies y=\frac{3kx^{\frac{4}{3}}}{4}+C$$ Since, the curve passes through $(1, 4)$ hence satisfying the above equation by this point as follows $$4=\frac{3k(1)^{\frac{4}{3}}}{4}+C\implies 4=\frac{3k}{4}+C$$$$\implies C=\frac{16-3k}{4}\tag 1$$
Similarly, satisfying the above equation by the point $(8, 16)$ as follows $$16=\frac{3k(8)^{\frac{4}{3}}}{4}+C\implies 16=\frac{48k}{4}+C$$$$\implies C=16-12k\tag 2$$ Equating (1) & (2), we get
$$\frac{16-3k}{4}=16-12k$$ $$\implies k=\frac{48}{45}=\frac{16}{15}$$ $$\implies C=16-12\left(\frac{16}{15}\right)=\frac{80-64}{5}=\frac{16}{5}$$ Now, substituting the values of $k$ & $C$, we get the equation of the curve $$y=\frac{3\left(\frac{16}{15}\right)x^{\frac{4}{3}}}{4}+\frac{16}{5}$$  $$\color{blue}{y=\frac{4}{5}x^{\frac{4}{3}}+\frac{16}{5}}$$ 
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