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Example solving non-linear first order ODE

Nasser M. Abbasi

September 8, 2018   Compiled on May 18, 2020 at 6:46pm

\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*}