# how to solve and sketch $\int\frac{1}{y^{4/5}(1-y)}\,\mathrm{d}y=\,\int\mathrm{d}x$

I want to solve the following equation and then be able to sketch the curve for positive values of $x$: \begin{align} \int\frac{1}{y^{4/5}(1-y)}\,\mathrm{d}y&=\,\int\mathrm{d}x\\ &=\,x+c, \end{align} where $c$ is a constant of integration.

The thing is I don't know how to solve the integral; the wolfram engine gives a very long complicated answer which makes the sketching part quite cumbersome.

So, what would be an alternative and efficient way to solve this problem?

Is numerical integration appropriate here? Should I try to solve it using a programming language or maybe MATLAB?

For $y\approx0$ the differential equation looks like $y'=y^{4/5}$ which is separable and solves as $(y^{1/5})'=\frac15y^{-4/5}y'=\frac15$ or $$y(t)=\Bigl(y_0^{1/5}+\tfrac15(t-t_0)\Bigr)^5.$$ This means solutions move away from $0$ with increasing $t$. $y\equiv 0$ is a solution and solutions can branch out at any time, $y(t)=0$ for $t<t_0$, $y(t)=\Bigl(\frac{t-t_0}5\Bigr)^5$ for $t\ge t_0$.

For $y\approx 1$ the DE is close to $y'=1-y$ or $$y=1+(y_0-1)e^{t_0-t}.$$ This means that $y\equiv 1$ is a stable stationary solution.

In between the solutions are monotonically increasing for $y\in(0,1)$ and decreasing for $y>1$. This should be sufficient for a sketch of the solution family.

To get an exact solution use the substitution $y=u^5$, $dy=5u^4du$ to transform $$\int \frac{dy}{y^{4/5}(1-u)}=\int\frac{5du}{1-u^5}$$ This can be solved via partial fraction decomposition and leads to the observed terms of high complexity.

Do you want to have the graph or the inverse function ?

The graph you can get as follows, there is no need for an inversion:

First sketch the curve for $\,\displaystyle y=-c+\int\frac{1}{x^{4/5}(1-x)}dx\,$ and then reflect the graph

at the bisector of the first and third part (means: at the line $\,y=x\,$) .

Note:

Wolfram alpha gives the approximation

$\displaystyle\int\frac{1}{x^{4/5}(1-x)}=-1.90211\arctan(0.32492-1.05146 \sqrt[5]{x} )-$

$-0.309017\ln(\sqrt[5]{x^2}-0.618034\sqrt[5]{x}+1)+0.809017\ln(\sqrt[5]{x^2}+1.61803\sqrt[5]{x}+1)-$

$-\ln(1-\sqrt[5]{x})+1.17557\arctan(1.7013\sqrt[5]{x}+1.37638) + constant$

which seems to be good enough for a calculation to get the graph.

The exact solution is:

$\displaystyle\int\frac{1}{x^{4/5}(1-x)}=$

$\displaystyle-\frac{\sqrt{5}-1}{4} \ln(\sqrt[5]{x^2}-\frac{\sqrt{5}-1}{2}\sqrt[5]{x}+1)+\frac{\sqrt{5}+1}{4} \ln(\sqrt[5]{x^2}+\frac{\sqrt{5}+1}{2}\sqrt[5]{x}+1)-\ln(1-\sqrt[5]{x})$

$\displaystyle+\frac{1}{2}\sqrt{10+2\sqrt{5}} \arctan(\frac{4\sqrt[5]{x}-\sqrt{5}+1}{\sqrt{10+2\sqrt{5}}})+ \frac{1}{2}\sqrt{10-2\sqrt{5}}\arctan(\frac{4\sqrt[5]{x}+\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}})$

$+ constant$

• Thank you for your answer. I am not sure how to actually sketch the curve. Obviously I need to solve the integral, right? – johnny09 Aug 11 '17 at 10:11
• Yes.:-) Special case: For $|x|<1$ you can use the integral of the series expansion for an approximation. Use the transformation of LutzL (see his answer). – user90369 Aug 11 '17 at 10:26