If $\frac{dy}{dx} = y^2$, show that $y =\frac{ -1}{x + C}$. I tried integrating both sides but it doesn't seem to be working :l
If $\frac{dy}{dx} = y^2$, show that $y =\frac{ -1}{x + C}$
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ordinary-differential-equations
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1**Hint:** $\displaystyle \int \dfrac{dy}{y^2} = \int dx$. In other words, it is a Separable Equation. – 2017-02-03
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0Thank you very much! ^-^ all solved – 2017-02-03
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0If the question goes like "show that....." then I think it suffices to plug in the suggested solution into the DFQ and show it becomes an identity. Nothing wrong with actually solving the DFQ, but sometimes that could be more work. – 2017-02-03
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0I wanted to know how I'd get there without using Wolfram :) – 2017-02-03
3 Answers
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This is a seperable differential equation. This should get you started:$$\begin{align}\frac{dy}{dx}&=y^2\\\frac{dy}{y^2}&=dx\\\int\frac{1}{y^2}dy&=\int dx+C\end{align}$$ Then integrate both sides, and rearrange for $y$.
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Integrating both sides does not work, since the RHS is a function of $y$ which is itself a function of $x$. So integrating w.r.t. to $x$ does not lead you anywhere. Remember, you want to find the function $y(x)$, not some function of $y$. So as the previous answer suggests, sperarate, i.e. "heuristically speaking" bring the differentials $dx$ everand $dy$ and the respective expressions of $x$ and $y$ on each side:
$$\frac{1}{y^2} dy = dx$$
Now you can integrate each side with respect to their respective variables ($y$ and $x$). There is also another condition, called initial value, i.e. the function $y(x)$ at some value, most often $0$, so $y_0 = y(0)$. Now
$$\int_{y_0}^y \frac{1}{y^2} dy = -\left(\frac{1}{y} - \frac{1}{y_0} \right)$$
and
$$\int_0^x x'dx' = x $$
Therefore:
$$-\left(\frac{1}{y} - \frac{1}{y_0}\right) = x $$
and solving for $y$ yields:
$$y = \frac{1}{\frac{1}{y_0} - x} $$
And you see that the initial condition automatically defines your constant $C$.
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$$f'(x)=f(x)^2$$ $$\frac{f'(x)}{f(x)^2}=1$$ $$\int\frac{f'(x)}{f(x)^2}dx=\int dx$$ Let $y=f(x)$. Thus, $dy=f'(x)dx$. $$\int\frac{dy}{y^2}=x+C$$ $$\frac{y^{1-2}}{1-2}=x+C$$ $$\frac{-1}{f(x)}=x+C$$ $$-f(x)=\frac{1}{x+C}$$ $$f(x)=\frac{-1}{x+C}$$ And there you go.