ode internal name "second_order_ode_missing_x"
Given \begin {equation} y^{\prime \prime }+f\left ( y^{\prime }\right ) +g\left ( y\right ) =0 \tag {1} \end {equation} Let \(p=y^{\prime }\) then \(y^{\prime \prime }=\frac {dp}{dx}=\frac {dp}{dy}\frac {dy}{dx}=\frac {dp}{dy}p\) and the ode becomes\begin {equation} p\frac {dp}{dy}+f\left ( p\right ) +g\left ( y\right ) =0 \tag {2} \end {equation} Which is now a first order ode.
3.4.6.1.1 Example 1 \[ yy^{\prime \prime }-\left ( y^{\prime }\right ) ^{2}=1 \] Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes\begin {align*} yp\frac {dp}{dy}-p^{2} & =1\\ \frac {dp}{dy} & =\frac {1+p^{2}}{p}\frac {1}{y} \end {align*}
This is separable. \begin {align*} \frac {dp}{dy}\frac {p}{1+p^{2}} & =\frac {1}{y}\\ \frac {p}{1+p^{2}}dp & =\frac {1}{y}dy\\ \int \frac {p}{1+p^{2}}dp & =\int \frac {1}{y}dy\\ \frac {1}{2}\ln \left ( p-1\right ) +\frac {1}{2}\ln \left ( p+1\right ) & =\ln y+c \end {align*}
Or, assuming \(p-1>0,p+1>0\)\begin {align*} \ln \left ( p-1\right ) +\ln \left ( p+1\right ) & =2\ln y+2c\\ \ln \left ( \left ( p-1\right ) \left ( p+1\right ) \right ) & =\ln y^{2}+c_{1}\\ \left ( p-1\right ) \left ( p+1\right ) & =c_{2}y^{2}\\ p^{2}-1 & =c_{2}y^{2}\\ p^{2} & =c_{2}y^{2}+1 \end {align*}
Hence\[ p=\pm \sqrt {1+c_{2}y^{2}}\] But \(p=y^{\prime }\left ( x\right ) \). The above becomes\[ y^{\prime }\left ( x\right ) =\pm \sqrt {1+c_{2}y^{2}}\] This is first order ode which is separable. The first one gives\begin {align*} y^{\prime }\left ( x\right ) & =\sqrt {1+c_{2}y^{2}}\\ \frac {dy}{\sqrt {1+c_{2}y^{2}}} & =dx\\ \int \frac {dy}{\sqrt {1+c_{2}y^{2}}} & =\int dx\\ \frac {1}{\sqrt {c_{2}}}\ln \left ( \sqrt {c_{2}}y+\sqrt {1+c_{2}y^{2}}\right ) & =x+c_{3}\\ \ln \left ( \sqrt {c_{2}}y+\sqrt {1+c_{2}y^{2}}\right ) & =\sqrt {c_{2}}x+\sqrt {c_{2}}c_{3} \end {align*}
Where \(c_{2},c_{3}\) are constants. Similar solution result for the negative ode.
3.4.6.1.2 Example 2 \begin {equation} y^{\prime \prime }+ay\left ( y^{\prime }\right ) +by^{3}=0 \tag {1} \end {equation} Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes\begin {equation} p\frac {dp}{dy}+ayp+by^{3}=0 \tag {2} \end {equation} Which is now a first order ode. \begin {equation} \frac {dp}{dy}=-ay+b\frac {y^{3}}{p} \tag {3} \end {equation} Solving for \(p\) gives\[ \frac {1}{4\sqrt {a^{2}+8b}}\left ( \ln \left ( -by^{4}+ay^{2}p+2p^{2}\right ) \sqrt {a^{2}+8b}+2a\operatorname {arctanh}\left ( \frac {ax^{2}+4p}{y^{2}\sqrt {a^{2}+8b}}\right ) \right ) =c_{1}\] Then \(y\) is found by solving \(y^{\prime }=p\), another first order ode.\[ \frac {1}{4\sqrt {a^{2}+8b}}\left ( \ln \left ( -by^{4}+ay^{2}y^{\prime }+2\left ( y^{\prime }\right ) ^{2}\right ) \sqrt {a^{2}+8b}+2a\operatorname {arctanh}\left ( \frac {ax^{2}+4y^{\prime }}{y^{2}\sqrt {a^{2}+8b}}\right ) \right ) =c_{1}\] But this second one could not solve. Actually ode (3) is homogeneous, class G and should use formula given in Kamke’s book, p. 19. but I have yet to implement this.
3.4.6.1.3 Example 3 \begin {equation} 2yy^{\prime \prime }-y^{3}-2\left ( y^{\prime }\right ) ^{2}=0 \tag {1} \end {equation} With IC\begin {align*} y\left ( 0\right ) & =-1\\ y^{\prime }\left ( 0\right ) & =0 \end {align*} Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes\begin {align} 2yp\frac {dp}{dy}-y^{3}-2p^{2} & =0\tag {2}\\ \frac {dp}{dy} & =\frac {y^{3}+2p^{2}}{2py}\nonumber \end {align}
Which is first order ode in \(p\left ( y\right ) \) of type Bernoulli. There are two solutions\begin {align} p_{1} & =y\sqrt {y+c_{1}}\tag {3}\\ p_{2} & =-y\sqrt {y+c_{1}} \tag {4} \end {align}
But \(p=y^{\prime }\) hence the above becomes\begin {align} y^{\prime }\left ( x\right ) & =y\sqrt {y+c_{1}}\tag {3}\\ y^{\prime }\left ( x\right ) & =-y\sqrt {y+c_{1}} \tag {4} \end {align}
Solving (3). At \(x=0\) we have \(y^{\prime }\left ( 0\right ) =0,y\left ( 0\right ) =-1\) hence the above becomes
\begin {align*} 0 & =-1\sqrt {-1+c_{1}}\\ 0 & =\sqrt {-1+c_{1}}\\ c_{1} & =1 \end {align*}
Hence (3) becomes
\[ y^{\prime }\left ( x\right ) =y\sqrt {y+1}\]
This is quadrature. Integrating
\begin {align*} \frac {dy}{y\sqrt {y+1}} & =dx\\ -2\operatorname {arctanh}\left ( \sqrt {y+1}\right ) & =x+c_{2} \end {align*}
At \(x=0\) we have\(\ y\left ( 0\right ) =-1\) and the above becomes
\begin {align*} -2\operatorname {arctanh}\left ( \sqrt {-1+1}\right ) & =c_{2}\\ c_{2} & =-2\operatorname {arctanh}\left ( 0\right ) \\ c_{2} & =0 \end {align*}
Hence the solution is
\begin {align*} -2\operatorname {arctanh}\left ( \sqrt {y+1}\right ) & =x\\ \operatorname {arctanh}\left ( \sqrt {y+1}\right ) & =-\frac {x}{2}\\ \sqrt {y+1} & =\tanh \left ( -\frac {x}{2}\right ) \\ & =-\tanh \left ( \frac {x}{2}\right ) \\ y+1 & =\tanh ^{2}\left ( \frac {x}{2}\right ) \\ y & =\tanh ^{2}\left ( \frac {x}{2}\right ) -1 \end {align*}
Solving (4)
At \(x=0\) we have \(y^{\prime }\left ( 0\right ) =0,y\left ( 0\right ) =-1\) hence
\begin {align*} 0 & =1\sqrt {-1+c_{1}}\\ 0 & =\sqrt {-1+c_{1}}\\ c_{1} & =1 \end {align*}
Hence (4) becomes
\[ y^{\prime }\left ( x\right ) =y\sqrt {y+1}\]
Which gives same solution as before. \(y=\tanh ^{2}\left ( \frac {x}{2}\right ) -1\)
3.4.6.1.4 Example 4 \begin {equation} 2y^{\prime \prime }-e^{y}=0 \tag {1} \end {equation} With IC\begin {align*} y\left ( 0\right ) & =0\\ y^{\prime }\left ( 0\right ) & =1 \end {align*} Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes\begin {align} 2p\frac {dp}{dy}-e^{y} & =0\nonumber \\ 2\frac {dp}{dy}p & =e^{y} \tag {2} \end {align}
This is separable. \begin {align} 2\int pdp & =\int e^{y}dy\nonumber \\ p^{2} & =e^{y}+c_{1} \tag {3} \end {align}
Before solving this, we should apply IC now as it simplifies the solution greatly. This assumes both \(y,y^{\prime }\) are are given at same point \(x_{0}\). Which is the case here. If only one IC is given (such as \(y\left ( 0\right ) \) or \(y^{\prime }\left ( 0\right ) \) but not both, then we can not apply IC now and have to do it at the end).
We are given that \(y^{\prime }\left ( 0\right ) =p=1,y\left ( 0\right ) =0\), hence the above reduces to\begin {align*} 1 & =e^{0}+c_{1}\\ c_{1} & =0 \end {align*}
Hence (3) now becomes\[ p^{2}=e^{y}\] but \(p=y^{\prime }\) hence\begin {align*} \left ( y^{\prime }\right ) ^{2} & =e^{y}\\ y^{\prime } & =\pm \sqrt {e^{y}} \end {align*}
This is quadrature. For the positive solution\begin {align} \frac {dy}{\sqrt {e^{y}}} & =dx\tag {4}\\ \frac {2}{\sqrt {e^{y}}} & =-x+c_{2} \end {align}
For \(y\left ( 0\right ) =0\) we obtain\[ 2=c_{2}\] Hence (4) becomes\begin {align*} \frac {2}{\sqrt {e^{y}}} & =-x+2\\ \sqrt {e^{y}} & =\frac {2}{2-x}\\ e^{y} & =\left ( \frac {2}{2-x}\right ) ^{2}\\ y_{1} & =2\ln \left ( \frac {2}{2-x}\right ) \end {align*}
For the negative solution\[ y^{\prime }=-\sqrt {e^{y}}\] Integrating\begin {equation} \frac {2}{\sqrt {e^{y}}}=x+c_{2} \tag {5} \end {equation} At \(y\left ( 0\right ) =0\)\[ 2=c_{2}\] Hence (5) becomes\begin {align*} \frac {2}{\sqrt {e^{y}}} & =x+2\\ \sqrt {e^{y}} & =\frac {2}{x+2}\\ e^{y} & =\left ( \frac {2}{x+2}\right ) ^{2}\\ y_{2} & =2\ln \left ( \frac {2}{x+2}\right ) \end {align*}
However, this solution do not satisfy \(y^{\prime }\left ( 0\right ) =1\) so it is discarded. Hence the solution is only\[ y_{1}=2\ln \left ( \frac {2}{2-x}\right ) \]
3.4.6.1.5 Example 5 This is same example as above, but here we delay applying IC to the very end to see the difference. This method is more general, but makes solving for IC harder.\begin {equation} 2y^{\prime \prime }-e^{y}=0 \tag {1} \end {equation} With IC\begin {align*} y\left ( 0\right ) & =0\\ y^{\prime }\left ( 0\right ) & =1 \end {align*}
Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes\begin {equation} 2\frac {dp}{dy}p=e^{y}\nonumber \end {equation} This is separable. \begin {align*} 2\int pdp & =\int e^{y}dy\\ p^{2} & =e^{y}+c_{1} \end {align*}
but \(p=y^{\prime }\) hence the above becomes\begin {align*} \left ( y^{\prime }\right ) ^{2} & =e^{y}+c_{1}\\ y^{\prime } & =\pm \sqrt {e^{y}+c_{1}} \end {align*}
This is quadrature. For the positive solution\begin {align} \frac {dy}{\sqrt {e^{y}+c_{1}}} & =dx\nonumber \\ -\frac {2\operatorname {arctanh}\left ( \frac {\sqrt {e^{y}+c_{1}}}{\sqrt {c_{1}}}\right ) }{\sqrt {c_{1}}} & =x+c_{2}\nonumber \\ 2\operatorname {arctanh}\left ( \frac {\sqrt {e^{y}+c_{1}}}{\sqrt {c_{1}}}\right ) & =-x\sqrt {c_{1}}-c_{2}\sqrt {c_{1}}\nonumber \\ \operatorname {arctanh}\left ( \frac {\sqrt {e^{y}+c_{1}}}{\sqrt {c_{1}}}\right ) & =-x\frac {\sqrt {c_{1}}}{2}-\frac {c_{2}\sqrt {c_{1}}}{2}\nonumber \\ \frac {\sqrt {e^{y}+c_{1}}}{\sqrt {c_{1}}} & =\tanh \left ( -x\frac {\sqrt {c_{1}}}{2}-\frac {c_{2}\sqrt {c_{1}}}{2}\right ) \nonumber \\ \sqrt {e^{y}+c_{1}} & =\sqrt {c_{1}}\tanh \left ( -x\frac {\sqrt {c_{1}}}{2}-\frac {c_{2}\sqrt {c_{1}}}{2}\right ) \nonumber \\ e^{y}+c_{1} & =\left ( \sqrt {c_{1}}\tanh \left ( -x\frac {\sqrt {c_{1}}}{2}-\frac {c_{2}\sqrt {c_{1}}}{2}\right ) \right ) ^{2}\nonumber \\ e^{y} & =\left ( \sqrt {c_{1}}\tanh \left ( -x\frac {\sqrt {c_{1}}}{2}-\frac {c_{2}\sqrt {c_{1}}}{2}\right ) \right ) ^{2}-c_{1}\nonumber \\ y & =\ln \left ( \left ( \sqrt {c_{1}}\tanh \left ( -x\frac {\sqrt {c_{1}}}{2}-\frac {c_{2}\sqrt {c_{1}}}{2}\right ) \right ) ^{2}-c_{1}\right ) \tag {2} \end {align}
Now we have to use (2) and take derivative and solve for \(c_{1},c_{2}\). Much harder than if we have applied IC to each solution earlier.
3.4.6.1.6 Example 6 \begin {equation} 2y^{\prime \prime }-\sin \left ( 2y\right ) =0 \tag {1} \end {equation} With IC\begin {align*} y\left ( 0\right ) & =-\frac {\pi }{2}\\ y^{\prime }\left ( 0\right ) & =1 \end {align*} Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes\begin {align} 2p\frac {dp}{dy} & =\sin \left ( 2y\right ) \tag {2}\\ 2pdp & =\sin \left ( 2y\right ) dy\nonumber \\ \int 2pdp & =\int \sin \left ( 2y\right ) dy\nonumber \\ p^{2} & =-\frac {1}{2}\cos \left ( 2y\right ) +c_{1}\nonumber \end {align}
At \(x=0\) we have \(p=1,y=-\frac {\pi }{2}\). Hence the above becomes\begin {align*} 1 & =-\frac {1}{2}\cos \left ( -\pi \right ) +c_{1}\\ & =-\frac {1}{2}\cos \left ( \pi \right ) +c_{1}\\ 1 & =\frac {1}{2}+c_{1}\\ c_{1} & =\frac {1}{2} \end {align*}
Therefore (2) becomes\[ \left ( y^{\prime }\left ( x\right ) \right ) ^{2}=-\frac {1}{2}\cos \left ( 2y\right ) +\frac {1}{2}\] Need to solve and apply IC \(y\left ( 0\right ) =-\frac {\pi }{2}\) to finish.
3.4.6.1.7 Example 7 \begin {equation} yy^{\prime \prime }-\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{3}=0 \tag {1} \end {equation} With IC\begin {align*} y\left ( 0\right ) & =-1\\ y^{\prime }\left ( 0\right ) & =0 \end {align*} Let \(p=y^{\prime }\) then \(y^{\prime \prime }=\frac {dp}{dx}=\frac {dp}{dy}\frac {dy}{dx}=\frac {dp}{dy}p\). Hence the ode becomes\begin {align} y\frac {dp}{dy}p-p^{2}+p^{3} & =0\tag {2}\\ p^{\prime } & =\frac {p^{2}-p^{3}}{yp}\nonumber \\ p^{\prime } & =\frac {p-p^{2}}{y}\nonumber \end {align}
This is separable. Solving\[ \int \frac {dp}{p^{2}-p}=-\int \frac {1}{y}dy\qquad p-p^{2}\neq 0 \] This gives\[ \frac {p-1}{p}=\frac {c_{1}}{y}\] Applying IC \(p=0\) at \(y=-1\) show there is no solution as we obtain \(-1=0\). Hence no general solution exists. Let look for singular solution. This happens when \(p-p^{2}=0\) or \(p=0\) and \(p=1\). Looking at \(p=0\) means \(y^{\prime }=0\) or \(y=c\). At IC this gives \(c=-1\). Hence \(y=-1\). This also satisfies \(y^{\prime }\left ( 0\right ) =0\). So \(y=-1\) is valid singular solution. Let look at \(p=1\) which means \(y^{\prime }=1\) or \(y=x+c_{1}\). At first IC this gives \(c_{1}=-1\). Hence solution now becomes \(y=x-1\). But this does not satisfy \(y^{\prime }\left ( 0\right ) =0\). Therefore only \[ y=-1 \] Is solution (singular).