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## Example solving non-linear ﬁrst order ODE

September 8, 2018   Compiled on January 31, 2024 at 4:15am

\begin {align*} \frac {dy}{dt}+y^{\frac {3}{2}}\left ( t\right ) & =a^{\frac {3}{2}}\\ y\left ( 0\right ) & =0 \end {align*}

Write as

\begin {align} \left ( y^{\frac {3}{2}}-a^{\frac {3}{2}}\right ) dt+dy & =0\nonumber \\ M\left ( t,y\right ) dt+N\left ( t,y\right ) dy & =0\tag {1} \end {align}

Where

\begin {align*} M & =y^{\frac {3}{2}}-a^{\frac {3}{2}}\\ N & =1 \end {align*}

Check if exact

\begin {align*} \frac {\partial M\left ( t,y\right ) }{\partial y} & =\frac {3}{2}y^{\frac {1}{2}}\\ \frac {\partial N\left ( t,y\right ) }{\partial t} & =0 \end {align*}

Since $$\frac {\partial M\left ( t,y\right ) }{\partial y}\neq \frac {\partial N\left ( t,y\right ) }{\partial t}$$ 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 {align*} \mu & =e^{\int Ady}\\ & =e^{\int \frac {-\frac {3}{2}y^{\frac {1}{2}}}{y^{\frac {3}{2}}-a^{\frac {3}{2}}}dy}\\ & =e^{-\ln \left ( a^{\frac {3}{2}}-y^{\frac {3}{2}}\right ) }\\ & =\frac {1}{a^{\frac {3}{2}}-y^{\frac {3}{2}}} \end {align*}

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 {align} \frac {dU}{dt} & =\mu M=\frac {y^{\frac {3}{2}}-a^{\frac {3}{2}}}{a^{\frac {3}{2}}-y^{\frac {3}{2}}}=-1\tag {2}\\ \frac {dU}{dy} & =\mu N=\frac {1}{a^{\frac {3}{2}}-y^{\frac {3}{2}}}\tag {3} \end {align}

From (2)

\begin {align} U & =-\int dt\nonumber \\ & =-t+f\left ( y\right ) \tag {4} \end {align}

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

\begin {align*} f^{\prime }\left ( y\right ) & =\frac {1}{a^{\frac {3}{2}}-y^{\frac {3}{2}}}\\ 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 \end {align*}

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 {align*} \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\\ 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 ) \\ C & =\frac {2\sqrt {3}}{3\sqrt {a}}\frac {\pi }{6}\\ C & =\frac {\pi \sqrt {3}}{9\sqrt {a}} \end {align*}

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

\begin {align*} 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}}\\ 3t\sqrt {a}+2\sqrt {3}\arctan \left ( \frac {\sqrt {a}+2\sqrt {y}}{\sqrt {3}\sqrt {a}}\right ) +6\ln \left ( \sqrt {a}-\sqrt {y}\right ) -\ln \left ( a+\sqrt {ay}+y\right ) & =\frac {\pi \sqrt {3}}{3} \end {align*}