ode internal name "second_order_ode_missing_y"
Given\[ y^{\prime \prime }+F\left ( x\right ) f\left ( y^{\prime }\right ) +g\left ( x\right ) =0 \] Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\). Hence the ode becomes\[ p^{\prime }+F\left ( x\right ) f\left ( p\right ) +g\left ( x\right ) =0 \] Which is first order ode.
3.4.6.2.1 Example 1 \begin {equation} y^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}+y^{\prime }=0 \tag {1} \end {equation} Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\). Hence the ode becomes\begin {equation} p^{\prime }+p^{2}+p=0 \tag {2} \end {equation} Which is now a first order separable ode. Its solution can be easily found to be\[ p=\frac {1}{c_{1}e^{x}-1}\] Hence\[ y^{\prime }\left ( x\right ) =\frac {1}{c_{1}e^{x}-1}\] Which is now solved for \(y\left ( x\right ) \) as first order, which gives by integration\[ y=\ln \left ( c_{1}e^{x}-c_{2}+1\right ) -x \]
3.4.6.2.2 Example 2 \begin {align} y^{\prime \prime }+\left ( y^{\prime }\right ) ^{2} & =1\tag {1}\\ y\left ( 0\right ) & =0\nonumber \\ y^{\prime }\left ( 0\right ) & =1\nonumber \end {align} Let \(p\left ( x\right ) =y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\) and the ode becomes\begin {align*} p^{\prime }+p^{2} & =1\\ p^{\prime } & =1-p^{2}\\ \frac {dp}{dx} & =1-p^{2}\\ \int \frac {dp}{1-p^{2}} & =\int dx\\ \operatorname {arctanh}\left ( p\right ) & =x+c_{1}\\ p & =\tanh \left ( x+c_{1}\right ) \end {align*}
At \(x=0,p=1\) hence \begin {equation} 1=\tanh \left ( c_{1}\right ) \nonumber \end {equation} There is no solution. Hence no general solution exist. Now we look for singular solution. This happens when \(1-p^{2}=0\) or \(p^{2}=1\) or \(p=\pm 1\). For \(p=1\) this means \(y^{\prime }=1\) or \(y=x+c\) which at IC gives \(c=0\). Hence singular solution is \[ y=x \] This satisfies both IC’s. If we try \(p=-1\) it gives \(y=-x\) but this does not satisfy IC. So only solution is \(y=x\).