4.4.7.2 missing \(y\left ( x\right ) \) where \(x\) can also be missing or not

ode internal name "second_order_ode_missing_y"

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\). Hence the ode becomes

4.4.7.2.1 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

\[ y=\ln \left ( c_{1}e^{x}-c_{2}+1\right ) -x \]

4.4.7.2.2 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 \(y=x\).

4.4.7.2.3 Example 3A

\begin{align} y^{\prime \prime } & =\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\tag {1}\\ y\left ( 0\right ) & =1\nonumber \end{align}

Notice that only one IC is given. Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\). Hence the ode becomes

\begin{equation} p^{\prime }=\sqrt {1+p^{2}} \tag {2}\end{equation}

We can’t use IC on this ode, since the IC is only on \(y\) and not \(y^{\prime }\). Solving this as first order gives

\[ p\left ( x\right ) =\sinh \left ( x+c_{1}\right ) \]

But \(p=y^{\prime }\) hence the above becomes

\[ y^{\prime }\left ( x\right ) =\sinh \left ( x+c_{1}\right ) \]

Now we solve this using the IC \(y\left ( 0\right ) =1\).  Solving the above gives

\begin{equation} y=\cosh \left ( x+c_{1}\right ) +c_{2} \tag {3}\end{equation}

Applying IC, and now we need to be careful. We need to solve for \(c_{2}\) and not \(c_{1}\).

\begin{align*} 1 & =\cosh \left ( 0+c_{1}\right ) +c_{2}\\ c_{2} & =1-\cosh \left ( c_{1}\right ) \end{align*}

Hence (3) becomes

\[ y\left ( x\right ) =\cosh \left ( x+c_{1}\right ) +1-\cosh \left ( c_{1}\right ) \]

4.4.7.2.4 Example 3B

\begin{align} y^{\prime \prime } & =\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\tag {1}\\ y\left ( 0\right ) & =1\nonumber \end{align}

This is slightly alternative way to solving the ode. Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\). Hence the ode becomes

\begin{equation} p^{\prime }=\sqrt {1+p^{2}} \tag {2}\end{equation}

Solving this as first order gives

\[ p\left ( x\right ) =\sinh \left ( x+c_{1}\right ) \]

But \(p=y^{\prime }\) hence the above becomes

\[ y^{\prime }\left ( x\right ) =\sinh \left ( x+c_{1}\right ) \]

Integrating gives

\begin{align} y & =\int \sinh \left ( x+c_{1}\right ) dx+c_{2}\nonumber \\ & =\cosh \left ( x+c_{1}\right ) +c_{2} \tag {3}\end{align}

Now we need to apply IC’s to find \(c_{1},c_{2}.\) We only have one IC \(y\left ( 0\right ) =1\). Applying this to the above solution gives

\begin{align*} 1 & =\cosh \left ( c_{1}\right ) +c_{2}\\ c_{2} & =1-\cosh \left ( c_{1}\right ) \end{align*}

Hence (3) becomes

\[ y\left ( x\right ) =\cosh \left ( x+c_{1}\right ) +1-\cosh \left ( c_{1}\right ) \]

4.4.7.2.5 Example 4

\begin{align} y^{\prime \prime } & =\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\tag {1}\\ y^{\prime }\left ( 0\right ) & =1\nonumber \end{align}

Notice that only one IC is given. Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\). Hence the ode becomes

\begin{equation} p^{\prime }=\sqrt {1+p^{2}} \tag {2}\end{equation}

Now we can use the IC on this ode, since the IC is on \(y^{\prime }\). Solving this as first order gives

\[ p\left ( x\right ) =\sinh \left ( x+c_{1}\right ) \]

Applying IC, where \(p\left ( 0\right ) =y^{\prime }\left ( 0\right ) =1\) gives

\begin{align*} 1 & =\sinh \left ( c_{1}\right ) \\ c_{1} & =\operatorname {arcsinh}\left ( 1\right ) \end{align*}

Hence

\[ p\left ( x\right ) =\sinh \left ( x+\operatorname {arcsinh}\left ( 1\right ) \right ) \]

But \(p=y^{\prime }\) hence the above becomes

\[ y^{\prime }\left ( x\right ) =\sinh \left ( x+\operatorname {arcsinh}\left ( 1\right ) \right ) \]

Solving as first order ode gives

\[ y\left ( x\right ) =\cosh \left ( x+\ln \left ( 1+\sqrt {2}\right ) \right ) +c_{2}\]

4.4.7.2.6 Example 5

\begin{equation} y^{\prime \prime }=\left ( y^{\prime }\right ) ^{2}\cos x \tag {1}\end{equation}

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\). Hence the ode becomes

\begin{align*} p^{\prime } & =p^{2}\cos x\\ \int \frac {dp}{p^{2}} & =\int \cos xdx\\ -\frac {1}{p} & =\sin x+c_{1}\end{align*}

Hence \(p=\frac {-1}{\sin x+c_{1}}\). But \(p=y^{\prime }\left ( x\right ) \). Therefore

\begin{align*} y^{\prime }\left ( x\right ) & =\frac {-1}{\sin x+c_{1}}\\ \int dy & =-\int \frac {dx}{\sin x+c_{1}}\\ y & =\frac {-\arctan \left ( \frac {2c_{1}\tan \left ( \frac {x}{2}\right ) +2}{2\sqrt {c_{1}^{2}-1}}\right ) }{\sqrt {c_{1}^{2}-1}}+c_{2}\end{align*}

4.4.7.2.7 Example 6

\begin{align} y^{\prime \prime } & =-\frac {1}{2\left ( y^{\prime }\right ) ^{2}}\tag {1}\\ y\left ( 0\right ) & =1\nonumber \\ y^{\prime }\left ( 0\right ) & =-1\nonumber \end{align}

Notice that this is also missing \(x\) type second order ode. Now let \(p\left ( x\right ) =y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\) and the ode becomes

\[ p^{\prime }=-\frac {1}{2p^{2}}\]

Which is quadrature. The solution is

\[ p^{3}+\frac {3x}{2}=c_{1}\]

At \(x=0,p\left ( 0\right ) =-1\). Hence the above gives

\[ -1=c_{1}\]

And the solution becomes

\[ p^{3}+\frac {3x}{2}=-1 \]

But \(p=y^{\prime }\), hence the above becomes

\[ \left ( y^{\prime }\right ) ^{3}+\frac {3x}{2}=-1 \]

With IC \(y\left ( 0\right ) =1\). This is quadrature. Solving gives

\begin{align*} y_{1} & =-\frac {1}{16}i\left ( 3x+2\right ) \left ( -i+\sqrt {3}\right ) \left ( -12x-8\right ) ^{\frac {1}{3}}+c_{1}\\ y_{2} & =\frac {1}{16}i\left ( 3x+2\right ) \left ( i+\sqrt {3}\right ) \left ( -12x-8\right ) ^{\frac {1}{3}}+c_{1}\\ y_{3} & =\frac {1}{8}\left ( 3x+2\right ) \left ( -12x-8\right ) ^{\frac {1}{3}}+c_{1}\end{align*}

Applying IC to the above shows that only second solution satisfies the original initial conditions with \(c=\frac {3}{2}\). Hence solution is

\[ y_{2}=\frac {1}{16}\left ( 3x+2\right ) \left ( i\sqrt {3}-1\right ) \left ( -12x-8\right ) ^{\frac {1}{3}}+\frac {3}{2}\]

Another option when solving these types of odes is not to plugin the IC until the very end. Like this. Starting with

\[ p^{3}+\frac {3x}{2}=c_{1}\]

We do not resolve the \(c_{1}\). But keep it. Since \(p=y^{\prime }\), hence the above becomes

\[ \left ( y^{\prime }\right ) ^{3}+\frac {3x}{2}=c_{1}\]

This is quadrature. Solving gives

\begin{align*} y_{1} & =\frac {1}{16}i\left ( 3x-2c_{1}\right ) \left ( i-\sqrt {3}\right ) \left ( 8c_{1}-12x\right ) ^{\frac {1}{3}}+c_{2}\\ y_{2} & =\frac {1}{16}i\left ( 3x-2c_{1}\right ) \left ( i+\sqrt {3}\right ) \left ( 8c_{1}-12x\right ) ^{\frac {1}{3}}+c_{2}\\ y_{3} & =\frac {1}{8}\left ( 3x-2c_{1}\right ) \left ( 8c_{1}-12x\right ) ^{\frac {1}{3}}+c_{2}\end{align*}

And only now we solve for \(c_{1},c_{2}\) from both initial conditions. Assuming we made no mistake, then same result will come out.