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## Example solving non-linear ﬁrst order ODE

September 8, 2018 compiled on — Saturday September 08, 2018 at 05:23 AM

$\begin{array}{llll}\hfill \frac{dy}{dt}+{y}^{\frac{3}{2}}\left(t\right)& ={a}^{\frac{3}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y\left(0\right)& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Write as

$\begin{array}{llll}\hfill \left({y}^{\frac{3}{2}}-{a}^{\frac{3}{2}}\right)dt+dy& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill M\left(t,y\right)dt+N\left(t,y\right)dy& =0\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\end{array}$

Where

$\begin{array}{llll}\hfill M& ={y}^{\frac{3}{2}}-{a}^{\frac{3}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill N& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Check if exact

$\begin{array}{llll}\hfill \frac{\partial M\left(t,y\right)}{\partial y}& =\frac{3}{2}{y}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\partial N\left(t,y\right)}{\partial t}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Since $\frac{\partial M\left(t,y\right)}{\partial y}\ne \frac{\partial N\left(t,y\right)}{\partial t}$ then Not exact. Trying integrating factor $A=\frac{\frac{\partial N}{\partial t}-\frac{\partial M}{\partial y}}{M}=\frac{-\frac{3}{2}{y}^{\frac{1}{2}}}{{y}^{\frac{3}{2}}-{a}^{\frac{3}{2}}}$, Since it is a function of $y$ alone, then it (1) can be made exact. The integrating factor is

$\begin{array}{llll}\hfill \mu & ={e}^{\int Ady}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{\int \frac{-\frac{3}{2}{y}^{\frac{1}{2}}}{{y}^{\frac{3}{2}}-{a}^{\frac{3}{2}}}dy}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{-ln\left({a}^{\frac{3}{2}}-{y}^{\frac{3}{2}}\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{{a}^{\frac{3}{2}}-{y}^{\frac{3}{2}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Multiplying (1) by this integrating factor, now it becomes exact

$\mu M\left(t,y\right)dt+\mu N\left(t,y\right)dy=0$

Now we follow standard method for solving exact ODE. Let

$\begin{array}{lll}\hfill \frac{dU}{dt}& =\mu M=\frac{{y}^{\frac{3}{2}}-{a}^{\frac{3}{2}}}{{a}^{\frac{3}{2}}-{y}^{\frac{3}{2}}}=-1\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \frac{dU}{dy}& =\mu N=\frac{1}{{a}^{\frac{3}{2}}-{y}^{\frac{3}{2}}}\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\end{array}$

From (2)

$\begin{array}{llll}\hfill U& =-\int dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-t+f\left(y\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\end{array}$

Substituting this into (3) to solve for $f\left(y\right)$

$\begin{array}{llll}\hfill {f}^{\prime }\left(y\right)& =\frac{1}{{a}^{\frac{3}{2}}-{y}^{\frac{3}{2}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill f\left(y\right)& =\frac{-2\sqrt{3}}{3\sqrt{a}}arctan\left(\frac{1+2\sqrt{\frac{y}{a}}}{\sqrt{3}}\right)-\frac{2}{3\sqrt{a}}ln\left(\sqrt{a}-\sqrt{y}\right)+\frac{1}{3\sqrt{a}}ln\left(a+\sqrt{ay}+y\right)+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Hence the solution from (4) is

$U=-t+\frac{-2\sqrt{3}}{3\sqrt{a}}arctan\left(\frac{1+2\sqrt{\frac{y}{a}}}{\sqrt{3}}\right)-\frac{2}{3\sqrt{a}}ln\left(\sqrt{a}-\sqrt{y}\right)+\frac{1}{3\sqrt{a}}ln\left(a+\sqrt{ay}+y\right)+C$

But $\frac{dU}{dt}=0$, hence $U={C}_{1}$. Therefore, collecting constants into one, the solution is (implicit form)

$t+\frac{2\sqrt{3}}{3\sqrt{a}}arctan\left(\frac{1+2\sqrt{\frac{y}{a}}}{\sqrt{3}}\right)+\frac{2}{3\sqrt{a}}ln\left(\sqrt{a}-\sqrt{y}\right)-\frac{1}{3\sqrt{a}}ln\left(a+\sqrt{ay}+y\right)=C$

From initial conditions

$\begin{array}{llll}\hfill \frac{2\sqrt{3}}{3\sqrt{a}}arctan\left(\frac{1}{\sqrt{3}}\right)+\frac{2}{3\sqrt{a}}ln\left(\sqrt{a}\right)-\frac{1}{3\sqrt{a}}ln\left(a\right)& =C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill C& =\frac{2\sqrt{3}}{3\sqrt{a}}\frac{\pi }{6}+\frac{2}{3\sqrt{a}}ln\left(\sqrt{a}\right)-\frac{1}{3\sqrt{a}}ln\left(a\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill C& =\frac{2\sqrt{3}}{3\sqrt{a}}\frac{\pi }{6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill C& =\frac{\pi \sqrt{3}}{9\sqrt{a}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Hence ﬁnal solution for $y\left(t\right)$ in implicit form is

$\begin{array}{llll}\hfill t+\frac{2\sqrt{3}}{3\sqrt{a}}arctan\left(\frac{1+2\sqrt{\frac{y}{a}}}{\sqrt{3}}\right)+\frac{2}{3\sqrt{a}}ln\left(\sqrt{a}-\sqrt{y}\right)-\frac{1}{3\sqrt{a}}ln\left(a+\sqrt{ay}+y\right)& =\frac{\pi \sqrt{3}}{9\sqrt{a}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3t\sqrt{a}+2\sqrt{3}arctan\left(\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{3}\sqrt{a}}\right)+6ln\left(\sqrt{a}-\sqrt{y}\right)-ln\left(a+\sqrt{ay}+y\right)& =\frac{\pi \sqrt{3}}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$