How to solve the ode once it is determined it is exact

In the examples above we did not show how to obtain or ﬁnd the ﬁrst integral \(R\left ( x,y,y^{\prime }\right ) \). Given an ode \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =0\) which is determined to be exact as above, then how to solve it? This is done by ﬁrst ﬁnding the ﬁrst integral \(R\). We need to ﬁnd \(R\left ( x,y,y^{\prime }\right ) \) such that

Once \(R\) is found, then we need to solve the ﬁrst order ode \(R\left ( x,y,y^{\prime }\right ) =c\) where \(R\) is now one order less that \(F\) so it should be simpler to solve. This ode might require another integration factor to solve depending on what it type turns out to be.

This reduces the order of the ode from second to ﬁrst order (since \(R\) is ﬁrst order). To ﬁnd \(R\left ( x,y,y^{\prime }\right ) \) the ﬁrst step is to write the given ode in this form

\begin{equation} F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =f\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+g\left ( x,y,y^{\prime }\right ) \tag {1}\end{equation}

\begin{align} F & =\frac {d}{dx}R\left ( x,y,y^{\prime }\right ) \nonumber \\ & =\frac {\partial R}{\partial x}\frac {dx}{dx}+\frac {\partial R}{\partial y}\frac {dy}{dx}+\frac {\partial R}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\nonumber \\ & =R_{x}+R_{y}y^{\prime }+R_{y^{\prime }}y^{\prime \prime } \tag {1A}\end{align}

And since \(y^{\prime \prime }=\Phi \left ( x,y,y^{\prime }\right ) \) then the above can also be written as

The above is same as Eq (1B) in the introduction above. Comparing (1,1A) shows that

\begin{align} f & =R_{y^{\prime }}\tag {2}\\ g & =R_{x}+R_{y}y^{\prime } \tag {3}\end{align}

At this point it is easier to replace \(y^{\prime }\) by \(p\). The above becomes

\begin{align} f & =R_{p}\tag {2}\\ g & =R_{x}+R_{y}p \tag {3}\end{align}

Using (2,3) we are able to determine \(R\). Note that \(R\) must exist since we checked the ode is exact and hence must have a ﬁrst integral. This method similar to how we ﬁnd \(R\) for an exact ﬁrst order ode.

\begin{equation} R=\int fdp+\psi \left ( x,y\right ) \tag {4}\end{equation}

Where \(\psi \left ( x,y\right ) \) acts like an integration constant but since \(R\) depends on more than one variable, it is now an arbitrary function of the other variables \(x,y\). If we can ﬁnd \(\psi \left ( x,y\right ) \), then \(R\) is found, since \(f\) is known. To ﬁnd \(\psi \,\), we diﬀerentiate one time w.r.t. \(x\) and another time w.r.t. \(y\) and substitute the result in (3). This gives

\begin{equation} g=\left ( \frac {\partial }{\partial x}\left ( \int fdp\right ) +\psi _{x}\left ( x,y\right ) \right ) +\left ( \frac {\partial }{\partial y}\left ( \int fdp\right ) +\psi _{y}\left ( x,y\right ) \right ) p \tag {5}\end{equation}

In the above the terms \(\frac {\partial }{\partial x}\left ( \int fdp\right ) ,\frac {\partial }{\partial y}\left ( \int fdp\right ) \) are known, since everything is known. The only unknowns are \(\psi _{x}\left ( x,y\right ) ,\psi _{y}\left ( x,y\right ) \). Comparing terms in (5) we can generate two equations for \(\psi _{x},\psi _{y}\) and by integrating them we ﬁnd \(\psi \). Examples below show how to do this as this is easier explained using examples.

4.4.1.4.1 Examples ﬁnding ﬁrst integral \(R\left ( x,y,y^{\prime }\right ) \) for an exact second order ode

Comparing this to \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =f\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+g\left ( x,y,y^{\prime }\right ) \) shows that

\begin{align*} f & =y\\ g & =\left ( y^{\prime }\right ) ^{2}+2axyy^{\prime }+ay^{2}\\ & =p^{2}+2axyp+ay^{2}\end{align*}

\begin{align} R & =\int fdp+\psi \left ( x,y\right ) \nonumber \\ & =yp+\psi \left ( x,y\right ) \tag {1A}\end{align}

But \(\frac {\partial }{\partial x}\left ( yp\right ) =0\) since \(y,p\) are held constant. It is important to watch for this here. Given \(f\left ( x,y\right ) =3x+y\left ( x\right ) \) where \(y\) is function of \(x\), then when we do \(\frac {\partial f}{\partial x}\) the result is \(3\) and not \(3+y^{\prime }\) because with partial derivatives the \(y\) is held constant. Similarly \(\frac {\partial }{\partial y}\left ( yp\right ) =p^{2}\). The above becomes

\begin{align*} p^{2}+2axyp+ay^{2} & =\psi _{x}+\left ( p+\psi _{y}\right ) p\\ & =\psi _{x}+p^{2}+\psi _{y}p\\ 2axyp+ay^{2} & =\psi _{x}+\psi _{y}p \end{align*}

\begin{align} 2axy & =\psi _{y}\tag {2A}\\ ay^{2} & =\psi _{x} \tag {3A}\end{align}

\begin{equation} \psi =axy^{2}+h\left ( x\right ) \tag {4A}\end{equation}

Diﬀerentiating the above w.r.t. \(x\) gives \(\psi _{x}=ay^{2}+h^{\prime }\left ( x\right ) \). comparing this to (3A) above gives \(ay^{2}=ay^{2}+h^{\prime }\left ( x\right ) \), hence \(h^{\prime }\left ( x\right ) =0\) or \(h\left ( x\right ) =c\). Therefore (4A) becomes

Therefore, since \(R=c_{1}\) a constant, then the above becomes (by merging the constants)

\begin{align*} yp+axy^{2} & =c_{2}\\ yy^{\prime }+axy^{2} & =c_{2}\end{align*}

This is the reduced ode which needs to be solved for \(y\). The above says that \(R=yy^{\prime }+axy^{2}+c_{2}\). To verify, let us apply \(F=\frac {d}{dx}R\). This gives

\begin{align*} yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}+2axyy^{\prime }+ay^{2} & =\frac {d}{dx}\left ( yy^{\prime }+axy^{2}+c_{2}\right ) \\ & =y^{\prime }y^{\prime }+yy^{\prime \prime }+ay^{2}+2axyy^{\prime }\\ & =yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}+2axyy^{\prime }+ay^{2}\end{align*}

\begin{align*} y^{\prime \prime }+xy^{\prime }+y & =0\\ F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) & =0 \end{align*}

This ode is not nonlinear, but let us apply this method to it anyway. First we need to determine if it is exact or not. Applying the test

\begin{align*} \frac {\partial F}{\partial y}-\frac {d}{dx}\left ( \frac {\partial F}{\partial y^{\prime }}\right ) +\frac {d^{2}}{dx^{2}}\left ( \frac {\partial F}{\partial y^{\prime \prime }}\right ) & =0\\ 1-\frac {d}{dx}\left ( x\right ) +\frac {d^{2}}{dx^{2}}\left ( 1\right ) & =0\\ 1-1 & =0\\ 0 & =0 \end{align*}

So it exact. Comparing this ode to \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =f\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+g\left ( x,y,y^{\prime }\right ) \) shows that

\begin{align*} f & =1\\ g & =xy^{\prime }+y\\ & =xp+y \end{align*}

\begin{align} R & =\int fdp+\psi \left ( x,y\right ) \nonumber \\ & =p+\psi \left ( x,y\right ) \tag {1A}\end{align}

But \(\frac {\partial p}{\partial x}=0\) since \(y\) is held constant. And \(\frac {\partial p}{\partial y}=0\). The above becomes

\begin{align*} x & =\psi _{y}\\ y & =\psi _{x}\end{align*}

Therefore, since \(R=c_{1}\) a constant, then the above becomes (by merging the constants)

\begin{align*} p+xy & =c_{2}\\ y^{\prime }+xy & =c_{2}\end{align*}

\[ y=\operatorname {erf}\left ( \frac {i\sqrt {2}x}{2}\right ) e^{\frac {-x^{2}}{2}}c_{1}+c_{2}e^{\frac {-x^{2}}{2}}\]

\begin{align*} y^{\prime \prime }-2yy^{\prime } & =0\\ F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) & =0 \end{align*}

\begin{align*} \frac {\partial F}{\partial y}-\frac {d}{dx}\left ( \frac {\partial F}{\partial y^{\prime }}\right ) +\frac {d^{2}}{dx^{2}}\left ( \frac {\partial F}{\partial y^{\prime \prime }}\right ) & =0\\ -2y^{\prime }-\frac {d}{dx}\left ( -2y\right ) +\frac {d^{2}}{dx^{2}}\left ( 1\right ) & =0\\ -2y^{\prime }+2\frac {d}{dx}\left ( y\right ) & =0\\ -2y^{\prime }+2y^{\prime } & =0\\ 0 & =0 \end{align*}

So it exact. Comparing this ode to \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =f\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+g\left ( x,y,y^{\prime }\right ) \) shows that

\begin{align*} f & =1\\ g & =-2yy^{\prime }\\ & =-2yp \end{align*}

\begin{align} R & =\int fdp+\psi \left ( x,y\right ) \nonumber \\ & =p+\psi \left ( x,y\right ) \tag {1A}\end{align}

\begin{align*} g & =\left ( \frac {\partial }{\partial x}\left ( \int fdp\right ) +\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}\left ( \int fdp\right ) +\psi _{y}\right ) p\\ -2yp & =\left ( \frac {\partial p}{\partial x}+\psi _{x}\right ) +\left ( \frac {\partial p}{\partial y}+\psi _{y}\right ) p\\ -2yp & =\psi _{x}+\psi _{y}p \end{align*}

\begin{align*} -2y & =\psi _{y}\\ 0 & =\psi _{x}\end{align*}

Integrating the ﬁrst equation gives \(\psi =-y^{2}+h\left ( x\right ) \). Diﬀerentiating this w.r.t. \(x\) gives \(\psi _{x}=h^{\prime }\left ( x\right ) \). comparing this to the second equation above gives \(0=h^{\prime }\left ( x\right ) \), hence \(h\left ( x\right ) =c\). Hence \(\psi =-y^{2}+c\). Therefore (1A) becomes

Therefore, since \(R=c_{1}\) a constant, then the above becomes (by merging the constants)

\begin{align*} p-y^{2} & =c_{2}\\ y^{\prime }-y^{2} & =c_{2}\end{align*}

\begin{align*} \left ( x-1\right ) ^{2}y^{\prime \prime }+4xy^{\prime }+2y-2x & =0\\ F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) & =0 \end{align*}

\begin{align*} \frac {\partial F}{\partial y}-\frac {d}{dx}\left ( \frac {\partial F}{\partial y^{\prime }}\right ) +\frac {d^{2}}{dx^{2}}\left ( \frac {\partial F}{\partial y^{\prime \prime }}\right ) & =0\\ 2-\frac {d}{dx}\left ( 4x\right ) +\frac {d^{2}}{dx^{2}}\left ( \left ( x-1\right ) ^{2}\right ) & =0\\ 2-4+\frac {d}{dx}\left ( 2\left ( x-1\right ) \right ) & =0\\ 2-4+2 & =0\\ 0 & =0 \end{align*}

So it exact. Comparing this ode to \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =f\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+g\left ( x,y,y^{\prime }\right ) \) shows that

\begin{align*} f & =\left ( x-1\right ) ^{2}\\ g & =4xy^{\prime }+2y-2x\\ & =4xp+2y-2x \end{align*}

\begin{align} R & =\int fdp+\psi \left ( x,y\right ) \nonumber \\ & =\left ( x-1\right ) ^{2}p+\psi \left ( x,y\right ) \tag {1A}\end{align}

\begin{align*} g & =\left ( \frac {\partial }{\partial x}\left ( \int fdp\right ) +\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}\left ( \int fdp\right ) +\psi _{y}\right ) p\\ 4xp+2y-2x & =\left ( \frac {\partial }{\partial x}\left ( \left ( x-1\right ) ^{2}p\right ) +\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}\left ( \left ( x-1\right ) ^{2}p\right ) +\psi _{y}\right ) p\\ 4xp+2y-2x & =2p\left ( x-1\right ) +\psi _{x}+\psi _{y}p\\ 4xp+2y-2x & =p\left ( 2\left ( x-1\right ) +\psi _{y}\right ) +\psi _{x}\end{align*}

\begin{align*} 4x & =2\left ( x-1\right ) +\psi _{y}\\ 2y-2x & =\psi _{x}\end{align*}

\begin{align*} 2x+2 & =\psi _{y}\\ 2y-2x & =\psi _{x}\end{align*}

Integrating the ﬁrst equation gives \(\psi =2xy+2y+h\left ( x\right ) \). Diﬀerentiating this w.r.t. \(x\) gives \(\psi _{x}=2y+h^{\prime }\left ( x\right ) \). comparing this to the second equation above gives \(2y-2x=2y+h^{\prime }\left ( x\right ) \), hence \(h^{\prime }\left ( x\right ) =-2x\). Hence \(h=-x^{2}+c\). Therefore\(\ \psi =2xy+2y-x^{2}+c\). Eq (1A) becomes

\begin{align*} R & =\left ( x-1\right ) ^{2}p+2xy+2y-x^{2}+c\\ & =\left ( x-1\right ) ^{2}y^{\prime }+2xy+2y-x^{2}+c \end{align*}

Therefore, since \(R=c_{1}\) a constant, then the above becomes (by merging the constants)

\begin{align*} y^{\prime \prime }-y^{\prime }e^{y} & =0\\ F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) & =0 \end{align*}

\begin{align*} \frac {\partial F}{\partial y}-\frac {d}{dx}\left ( \frac {\partial F}{\partial y^{\prime }}\right ) +\frac {d^{2}}{dx^{2}}\left ( \frac {\partial F}{\partial y^{\prime \prime }}\right ) & =0\\ -y^{\prime }e^{y}-\frac {d}{dx}\left ( -e^{y}\right ) +\frac {d^{2}}{dx^{2}}\left ( 1\right ) & =0\\ -y^{\prime }e^{y}+y^{\prime }e^{y} & =0\\ 0 & =0 \end{align*}

So it exact. Comparing this ode to \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =f\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+g\left ( x,y,y^{\prime }\right ) \) shows that

\begin{align*} f & =1\\ g & =-y^{\prime }e^{y}\\ & =-pe^{y}\end{align*}

\begin{align} R & =\int fdp+\psi \left ( x,y\right ) \nonumber \\ & =p+\psi \left ( x,y\right ) \tag {1A}\end{align}

\begin{align*} g & =\left ( \frac {\partial }{\partial x}\left ( \int fdp\right ) +\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}\left ( \int fdp\right ) +\psi _{y}\right ) p\\ -pe^{y} & =\left ( \frac {\partial }{\partial x}p+\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}p+\psi _{y}\right ) p\\ -pe^{y} & =\psi _{x}+\psi _{y}p \end{align*}

\begin{align*} -e^{y} & =\psi _{y}\\ 0 & =\psi _{x}\end{align*}

Integrating the ﬁrst equation gives \(\psi =-e^{y}+h\left ( x\right ) \). Partial diﬀerentiating this w.r.t. \(x\) gives \(\psi _{x}=h^{\prime }\left ( x\right ) \). comparing this to the second equation above gives \(h^{\prime }\left ( x\right ) =0\), hence \(h\left ( x\right ) =c\). Hence \(h=-x^{2}+c\). Therefore\(\ \psi =-e^{y}+c\). Eq (1A) becomes

\begin{align*} R & =p-e^{y}+c\\ & =y^{\prime }-e^{y}+c \end{align*}

Therefore, since \(\phi =c_{1}\) a constant, then the above becomes (by merging the constants)

4.4.1.4 How to solve the ode once it is determined it is exact