prirodnogbk

2022-07-07

The question asks to find y given y(x) is a differentiable function satisfying:
$\frac{dy}{dx}$=-$2x{y}^{4}$ , y(0)=$\frac{1}{3}$ and $y\left(x\right)>0$ , and explain why it is unique.
I assume the steps are to integrate $\frac{dy}{dx}$ to get a function for y(x), then use $y\left(0\right)=1/3$ to find the constant and then set $y\left(x\right)>0$ to find y but i keep getting stuck.
A differential equation solution gives y(x)=$\frac{1}{\sqrt[3]{{c}_{1}+3{x}^{2}}}$ , which I can use $y\left(0\right)$ to find ${c}_{1}$ but then I'm not sure how to get y from that since the equation doesn't have any y's in it.
A similar question I found on this site showed to rearrange and integrate each side like $\int \frac{dy}{{y}^{4}}=\int -2xdx$ which gives $-\frac{1}{3{y}^{3}}+{c}_{1}$=$-{x}^{2}+{c}_{2}$ , but then I don't know how to get $y\left(x\right)$ from that frunction, it seems further from the solution than my first try.

Kaylie Mcdonald

Expert

That's just a simple manipulation:
$\begin{array}{rl}-\frac{1}{3{y}^{3}}+{c}_{1}=-{x}^{2}+{c}_{2}& \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}-\frac{1}{3{y}^{3}}=-{x}^{2}+{c}_{2}-{c}_{1}\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3{y}^{3}}={x}^{2}+{c}_{1}-{c}_{2}\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{y}^{3}=\frac{1}{3{x}^{2}+3\left({c}_{1}-{c}_{2}\right)}\\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}y=\frac{1}{\sqrt[3]{3{x}^{2}+3\left({c}_{1}-{c}_{2}\right)}}.\end{array}$
Now it's just note that $3\left({c}_{1}-{c}_{2}\right)$ is constant, say $k$, and we obtain
$y=\frac{1}{\sqrt[3]{3{x}^{2}+k}}.$

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