4.4.7.1 Missing \(x\)

ode internal name "second_order_ode_missing_x"

Given

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

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=\frac {dp}{dx}=\frac {dp}{dy}\frac {dy}{dx}=\frac {dp}{dy}p=pp^{\prime }\) and the ode becomes

\begin{equation} pp^{\prime }=f\left ( y,p\right ) \tag {2}\end{equation}

Which is now a first order ode. If we can solve this for \(p\) then the solution to the original ode (1) is

\begin{align*} \frac {dy}{dx} & =p\left ( y\right ) \\ \int \frac {dy}{p\left ( y\right ) } & =x+c_{1}\end{align*}

4.4.7.1.1 Example 1

\[ yy^{\prime \prime }-\left ( y^{\prime }\right ) ^{2}=1 \]

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

\begin{align*} ypp^{\prime }-p^{2} & =1\\ p^{\prime } & =\frac {1+p^{2}}{p}\frac {1}{y}\end{align*}

This is separable.

\begin{align*} p^{\prime }\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

\[ p=\pm \sqrt {1+c_{2}y^{2}}\]

Therefore the solution to the original ode is

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

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.

4.4.7.1.2 Example 2

\begin{equation} y^{\prime \prime }+ay\left ( y^{\prime }\right ) +by^{3}=0 \tag {1}\end{equation}

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

\begin{equation} pp^{\prime }+ayp+by^{3}=0 \tag {2}\end{equation}

Which is now a first order ode.

\begin{equation} p^{\prime }=-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.

4.4.7.1.3 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 {3A}\\ y^{\prime }\left ( x\right ) & =-y\sqrt {y+c_{1}} \tag {4A}\end{align}

Before solving this ode, we can either use initial conditions to solve for \(c_{1}\) or solve it as it is and at the very end use initial conditions to solve for both \(c_{1}\) and the new constant which will come up which will be \(c_{2}\).  It is easier to get rid of \(c_{1}\) now than keep it. Will show both methods.

Getting rid of \(c_{1}\) now method. 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*}

Eq(3A) 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\nonumber \\ \operatorname {arctanh}\left ( \sqrt {y+1}\right ) & =-\frac {x}{2}\nonumber \\ \sqrt {y+1} & =\tanh \left ( -\frac {x}{2}\right ) \nonumber \\ & =-\tanh \left ( \frac {x}{2}\right ) \nonumber \\ y+1 & =\tanh ^{2}\left ( \frac {x}{2}\right ) \nonumber \\ y & =\tanh ^{2}\left ( \frac {x}{2}\right ) -1 \tag {5}\end{align}

Now we solve the second ode (4A). At \(x=0\) we have \(y^{\prime }\left ( 0\right ) =0,y\left ( 0\right ) =-1\) hence Eq.(4A) becomes

\begin{align*} 0 & =1\sqrt {-1+c_{1}}\\ 0 & =\sqrt {-1+c_{1}}\\ 1+c_{1} & =0\\ c_{1} & =-1 \end{align*}

Hence (4A) becomes

\[ y^{\prime }\left ( x\right ) =-y\sqrt {y-1}\]

Which gives the solution

\begin{equation} y\left ( x\right ) =x+2\arctan \left ( y-1\right ) +c_{2} \tag {6}\end{equation}

At \(x=0\) we have\(\ y\left ( 0\right ) =-1\) and the above becomes

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

Hence the solution (6) becomes

\[ y\left ( x\right ) =x+2\arctan \left ( y-1\right ) -1+2\arctan \left ( 2\right ) \]

But this solution does not satisfy \(y^{\prime }\left ( 0\right ) =0\). Hence it is not valid solution. So the only solution is (5).

Now we will do the same thing, but we will not get rid of \(c_{1}\) early one as above, and keep it until the end. We will see we will get same solution as (5).

Not getting rid of \(c_{1}\) method. Starting from (3A) and (4A) above.

\begin{align} y^{\prime }\left ( x\right ) & =y\sqrt {y+c_{1}}\tag {3A}\\ y^{\prime }\left ( x\right ) & =-y\sqrt {y+c_{1}} \tag {4A}\end{align}

Starting with (3A), solving it gives

\begin{align} \int \frac {1}{\sqrt {y+c_{1}}y}dy & =x+c_{2}\nonumber \\ -\frac {2\operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) }{\sqrt {c_{1}}} & =x+c_{2}\nonumber \\ -2\operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) & =x\sqrt {c_{1}}+c_{3}\nonumber \\ \operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) & =-x\frac {\sqrt {c_{1}}}{2}+c_{4}\nonumber \\ \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}} & =\tanh \left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ \sqrt {y+c_{1}} & =\sqrt {c_{1}}\tanh \left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ y+c_{1} & =c_{1}\tanh ^{2}\left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ y & =c_{1}\tanh ^{2}\left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) -c_{1} \tag {7}\end{align}

Now we can solve for the initial conditions. using \(y\left ( 0\right ) =-1\) gives

\begin{equation} -1=c_{1}\tanh ^{2}\left ( c_{4}\right ) -c_{1} \tag {8}\end{equation}

Taking derivative of the above gives

\[ y^{\prime }=-c_{1}^{\frac {3}{2}}\tanh \left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) \operatorname {sech}\left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) ^{2}\]

Applying \(y^{\prime }\left ( 0\right ) =0\) gives

\begin{equation} 0=-c_{1}^{\frac {3}{2}}\tanh \left ( c_{4}\right ) \operatorname {sech}\left ( c_{4}\right ) ^{2} \tag {9}\end{equation}

Solving (8,9) for \(c_{1},c_{4}\) gives

\begin{align*} c_{1} & =1\\ c_{4} & =0 \end{align*}

Hence the solution (7) is

\begin{equation} y=\tanh ^{2}\left ( -\frac {1}{2}x\right ) -1 \tag {10}\end{equation}

Which same as (5).  Now we go back and solve (4A).

\begin{align} y^{\prime }\left ( x\right ) & =-y\sqrt {y+c_{1}}\nonumber \\ \int \frac {1}{\sqrt {y+c_{1}}y}dy & =-x+c_{2}\\ -\frac {2\operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) }{\sqrt {c_{1}}} & =-x+c_{2}\nonumber \\ -2\operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) & =-x\sqrt {c_{1}}+c_{3}\nonumber \\ \operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) & =x\frac {\sqrt {c_{1}}}{2}+c_{4}\nonumber \\ \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}} & =\tanh \left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ \sqrt {y+c_{1}} & =\sqrt {c_{1}}\tanh \left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ y+c_{1} & =c_{1}\tanh ^{2}\left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ y & =c_{1}\tanh ^{2}\left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) -c_{1} \tag {7}\end{align}

Now we can solve for the initial conditions. using \(y\left ( 0\right ) =-1\) gives

\begin{equation} -1=c_{1}\tanh ^{2}\left ( c_{4}\right ) -c_{1} \tag {8}\end{equation}

Taking derivative of (7) gives

\[ y^{\prime }=c_{1}^{\frac {3}{2}}\tanh \left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) \left ( 1-\tanh \left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) ^{2}\right ) \]

Applying \(y^{\prime }\left ( 0\right ) =0\) gives

\begin{equation} 0=c_{1}^{\frac {3}{2}}\tanh \left ( c_{4}\right ) \left ( 1-\tanh \left ( c_{4}\right ) ^{2}\right ) \tag {9}\end{equation}

But now if we try to solve (8,9) for \(c_{1},c_{4}\) we see no solution exists. Hence (4A) leads to no solution. Only solution is (8). This is the same as earlier method.

This shows that if we get rid of \(c_{1}\) early one or not, same solution results. But it is much easier to get rid of \(c_{1}\) after finding the solution to the first ode.

4.4.7.1.4 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

\[ y_{1}=2\ln \left ( \frac {2}{2-x}\right ) \]

4.4.7.1.5 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.

4.4.7.1.6 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.

4.4.7.1.7 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).

4.4.7.1.8 Example 8

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

With IC

\begin{align*} y\left ( 0\right ) & =3\\ y^{\prime }\left ( 0\right ) & =\sqrt {2}\end{align*}

Let \(p=y^{\prime }\), hence \(y^{\prime \prime }=pp^{\prime }\) and the ode becomes

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

Solving the above ode gives

\begin{align} p_{1} & =-\frac {\sqrt {-2\left ( c_{1}y-1\right ) \left ( 2c_{1}y+y-2\right ) }}{2\left ( c_{1}y-1\right ) }\tag {3}\\ p_{2} & =\frac {\sqrt {-2\left ( c_{1}y-1\right ) \left ( 2c_{1}y+y-2\right ) }}{2\left ( c_{1}y-1\right ) } \tag {4}\end{align}

Now we replace back \(p=y^{\prime }\left ( x\right ) \) above gives

\begin{align} y^{\prime } & =-\frac {\sqrt {-2\left ( c_{1}y-1\right ) \left ( 2c_{1}y+y-2\right ) }}{2\left ( c_{1}y-1\right ) }\tag {3A}\\ y^{\prime } & =\frac {\sqrt {-2\left ( c_{1}y-1\right ) \left ( 2c_{1}y+y-2\right ) }}{2\left ( c_{1}y-1\right ) } \tag {4A}\end{align}

Lets start with (3A). Before solving, we will get rid of \(c_{1}\) using IC. Given that \(y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =\sqrt {2}\) then (3A) becomes

\begin{align*} \sqrt {2} & =-\frac {\sqrt {-2\left ( 3c_{1}-1\right ) \left ( 6c_{1}+3-2\right ) }}{2\left ( 3c_{1}-1\right ) }\\ c_{1} & =\frac {1}{6}\end{align*}

Hence (4A) becomes

\[ y^{\prime }=-\frac {\sqrt {-2\left ( \frac {1}{6}y-1\right ) \left ( 3y+y-2\right ) }}{2\left ( \frac {1}{6}y-1\right ) }\]

Solving this ode gives the solution

\begin{equation} x-\frac {\sqrt {-4y^{2}+30y-36}}{4}-\frac {9\arcsin \left ( \frac {4y}{9}-\frac {5}{3}\right ) }{8}+c_{2}=0 \tag {5}\end{equation}

Finally, using \(y\left ( 0\right ) =3\) the above becomes

\[ -\frac {\sqrt {-4\left ( 9\right ) +30\left ( 3\right ) -36}}{4}-\frac {9\arcsin \left ( \frac {4\left ( 3\right ) }{9}-\frac {5}{3}\right ) }{8}+c_{2}=0 \]

Solving for \(c_{2}\) gives

\[ c_{2}=\frac {\sqrt {18}}{4}-\frac {9\arcsin \left ( \frac {1}{3}\right ) }{8}\]

Hence (5) becomes

\begin{equation} x-\frac {\sqrt {-4y^{2}+30y-36}}{4}-\frac {9\arcsin \left ( \frac {4y}{9}-\frac {5}{3}\right ) }{8}+\frac {\sqrt {18}}{4}-\frac {9\arcsin \left ( \frac {1}{3}\right ) }{8}=0 \tag {6}\end{equation}

Now we have to do the same for ode (4A).  Given that \(y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =\sqrt {2}\) then (4A) becomes

\[ \sqrt {2}=\frac {\sqrt {-2\left ( 3c_{1}-1\right ) \left ( 6c_{1}+3-2\right ) }}{2\left ( 3c_{1}-1\right ) }\]

But there is no solution for \(c_{1}\). This means (4A) leads to no solution. Hence only solution is (6).