1.45 problem 45

Internal problem ID [8082]
Book : Second order enumerated odes
Section : section 1
Problem number : 45
Date solved : Monday, October 21, 2024 at 04:47:46 PM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} y {y^{\prime \prime }}^{2}+y^{\prime }&=0 \end{align*}

1.45.1 Solved as second order missing x ode

Time used: 10.010 (sec)

This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using

\begin{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} y p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

Solving for the derivative gives these ODE’s to solve

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

Now each of the above is solved separately.

Solving Eq. (1)

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(y,p) \mathop {\mathrm {d}y}+ N(y,p) \mathop {\mathrm {d}p}=0 \tag {1A} \]

Therefore

\begin{align*} \mathop {\mathrm {d}p} &= \left (-\frac {1}{\sqrt {-p y}}\right )\mathop {\mathrm {d}y}\\ \left (\frac {1}{\sqrt {-p y}}\right ) \mathop {\mathrm {d}y} + \mathop {\mathrm {d}p} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(y,p) &= \frac {1}{\sqrt {-p y}}\\ N(y,p) &= 1 \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial p} = \frac {\partial N}{\partial y} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial p} &= \frac {\partial }{\partial p} \left (\frac {1}{\sqrt {-p y}}\right )\\ &= -\frac {1}{2 \sqrt {-p y}\, p} \end{align*}

And

\begin{align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (1\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial y}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial y} \right ) \\ &=1\left ( \left ( \frac {y}{2 \left (-p y \right )^{{3}/{2}}}\right ) - \left (0 \right ) \right ) \\ &=-\frac {1}{2 \sqrt {-p y}\, p} \end{align*}

Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial y} - \frac {\partial M}{\partial p} \right ) \\ &=\sqrt {-p y}\left ( \left ( 0\right ) - \left (\frac {y}{2 \left (-p y \right )^{{3}/{2}}} \right ) \right ) \\ &=\frac {1}{2 p} \end{align*}

Since \(B\) does not depend on \(y\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then

\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {1}{2 p}\mathop {\mathrm {d}p} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{\frac {\ln \left (p \right )}{2} } \\ &= \sqrt {p} \end{align*}

\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= \sqrt {p}\left (\frac {1}{\sqrt {-p y}}\right ) \\ &= \frac {\sqrt {p}}{\sqrt {-p y}} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \sqrt {p}\left (1\right ) \\ &= \sqrt {p} \end{align*}

So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \\ \left (\frac {\sqrt {p}}{\sqrt {-p y}}\right ) + \left (\sqrt {p}\right ) \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (y,p\right )\)

\begin{align*} \frac {\partial \phi }{\partial y } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial p } &= \overline {N}\tag {2} \end{align*}

Integrating (2) w.r.t. \(p\) gives

\begin{align*} \int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \overline {N}\mathop {\mathrm {d}p} \\ \int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \sqrt {p}\mathop {\mathrm {d}p} \\ \tag{3} \phi &= \frac {2 p^{{3}/{2}}}{3}+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(y\) and \(p\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation}

But equation (1) says that \(\frac {\partial \phi }{\partial y} = \frac {\sqrt {p}}{\sqrt {-p y}}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {\sqrt {p}}{\sqrt {-p y}} = 0+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = \frac {\sqrt {p}}{\sqrt {-p y}} \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {\sqrt {p}}{\sqrt {-p y}}\right ) \mathop {\mathrm {d}y} \\ f(y) &= -\frac {2 \sqrt {-p y}}{\sqrt {p}}+ c_1 \\ \end{align*}

Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = \frac {2 p^{{3}/{2}}}{3}-\frac {2 \sqrt {-p y}}{\sqrt {p}}+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = \frac {2 p^{{3}/{2}}}{3}-\frac {2 \sqrt {-p y}}{\sqrt {p}} \]

Solving Eq. (2)

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(y,p) \mathop {\mathrm {d}y}+ N(y,p) \mathop {\mathrm {d}p}=0 \tag {1A} \]

Therefore

\begin{align*} \mathop {\mathrm {d}p} &= \left (\frac {1}{\sqrt {-p y}}\right )\mathop {\mathrm {d}y}\\ \left (-\frac {1}{\sqrt {-p y}}\right ) \mathop {\mathrm {d}y} + \mathop {\mathrm {d}p} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(y,p) &= -\frac {1}{\sqrt {-p y}}\\ N(y,p) &= 1 \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial p} = \frac {\partial N}{\partial y} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial p} &= \frac {\partial }{\partial p} \left (-\frac {1}{\sqrt {-p y}}\right )\\ &= \frac {1}{2 \sqrt {-p y}\, p} \end{align*}

And

\begin{align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (1\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial y}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial y} \right ) \\ &=1\left ( \left ( -\frac {y}{2 \left (-p y \right )^{{3}/{2}}}\right ) - \left (0 \right ) \right ) \\ &=\frac {1}{2 \sqrt {-p y}\, p} \end{align*}

Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial y} - \frac {\partial M}{\partial p} \right ) \\ &=-\sqrt {-p y}\left ( \left ( 0\right ) - \left (-\frac {y}{2 \left (-p y \right )^{{3}/{2}}} \right ) \right ) \\ &=\frac {1}{2 p} \end{align*}

Since \(B\) does not depend on \(y\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then

\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {1}{2 p}\mathop {\mathrm {d}p} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{\frac {\ln \left (p \right )}{2} } \\ &= \sqrt {p} \end{align*}

\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= \sqrt {p}\left (-\frac {1}{\sqrt {-p y}}\right ) \\ &= -\frac {\sqrt {p}}{\sqrt {-p y}} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \sqrt {p}\left (1\right ) \\ &= \sqrt {p} \end{align*}

So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \\ \left (-\frac {\sqrt {p}}{\sqrt {-p y}}\right ) + \left (\sqrt {p}\right ) \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (y,p\right )\)

\begin{align*} \frac {\partial \phi }{\partial y } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial p } &= \overline {N}\tag {2} \end{align*}

Integrating (2) w.r.t. \(p\) gives

\begin{align*} \int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \overline {N}\mathop {\mathrm {d}p} \\ \int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \sqrt {p}\mathop {\mathrm {d}p} \\ \tag{3} \phi &= \frac {2 p^{{3}/{2}}}{3}+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(y\) and \(p\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation}

But equation (1) says that \(\frac {\partial \phi }{\partial y} = -\frac {\sqrt {p}}{\sqrt {-p y}}\). Therefore equation (4) becomes

\begin{equation} \tag{5} -\frac {\sqrt {p}}{\sqrt {-p y}} = 0+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = -\frac {\sqrt {p}}{\sqrt {-p y}} \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -\frac {\sqrt {p}}{\sqrt {-p y}}\right ) \mathop {\mathrm {d}y} \\ f(y) &= \frac {2 \sqrt {-p y}}{\sqrt {p}}+ c_2 \\ \end{align*}

Where \(c_2\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = \frac {2 p^{{3}/{2}}}{3}+\frac {2 \sqrt {-p y}}{\sqrt {p}}+ c_2 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_2\) and \(c_2\) constants into the constant \(c_2\) gives the solution as

\[ c_2 = \frac {2 p^{{3}/{2}}}{3}+\frac {2 \sqrt {-p y}}{\sqrt {p}} \]

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} \frac {2 {y^{\prime }}^{{3}/{2}}}{3}-\frac {2 \sqrt {-y y^{\prime }}}{\sqrt {y^{\prime }}} = c_1 \end{align*}

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=\frac {\left (12 c_1 +24 \sqrt {-y}\right )^{{2}/{3}}}{4} \\ \tag{2} y^{\prime }&=\left (-\frac {\left (12 c_1 +24 \sqrt {-y}\right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (12 c_1 +24 \sqrt {-y}\right )^{{1}/{3}}}{4}\right )^{2} \\ \tag{3} y^{\prime }&=\left (-\frac {\left (12 c_1 +24 \sqrt {-y}\right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (12 c_1 +24 \sqrt {-y}\right )^{{1}/{3}}}{4}\right )^{2} \\ \tag{4} y^{\prime }&=\frac {\left (12 c_1 -24 \sqrt {-y}\right )^{{2}/{3}}}{4} \\ \tag{5} y^{\prime }&=\left (-\frac {\left (12 c_1 -24 \sqrt {-y}\right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (12 c_1 -24 \sqrt {-y}\right )^{{1}/{3}}}{4}\right )^{2} \\ \tag{6} y^{\prime }&=\left (-\frac {\left (12 c_1 -24 \sqrt {-y}\right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (12 c_1 -24 \sqrt {-y}\right )^{{1}/{3}}}{4}\right )^{2} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

\begin{align*} \int \frac {4}{\left (12 c_1 +24 \sqrt {-y}\right )^{{2}/{3}}}d y &= dx\\ -\frac {\left (-3 c_1 +2 \sqrt {-y}\right ) \left (12 c_1 +24 \sqrt {-y}\right )^{{1}/{3}}}{8}&= x +c_3 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_1 +24 \sqrt {-y}\right )^{{2}/{3}}}{4}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_1^{2}}{4} \end{align*}

Solving Eq. (2)

Integrating gives

\begin{align*} \int \frac {16}{\left (12 c_1 +24 \sqrt {-y}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d y &= dx\\ -\frac {\left (i \sqrt {3}-1\right ) \left (12 c_1 +24 \sqrt {-y}\right )^{{1}/{3}} \left (-3 c_1 +2 \sqrt {-y}\right )}{16}&= x +c_4 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_1 +24 \sqrt {-y}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}{16}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_1^{2}}{4} \end{align*}

Solving Eq. (3)

Integrating gives

\begin{align*} \int \frac {16}{\left (12 c_1 +24 \sqrt {-y}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d y &= dx\\ -\frac {3 \left (c_1 -\frac {2 \sqrt {-y}}{3}\right ) \left (1+i \sqrt {3}\right ) \left (12 c_1 +24 \sqrt {-y}\right )^{{1}/{3}}}{16}&= x +c_5 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_1 +24 \sqrt {-y}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}{16}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_1^{2}}{4} \end{align*}

Solving Eq. (4)

Integrating gives

\begin{align*} \int \frac {4}{\left (12 c_1 -24 \sqrt {-y}\right )^{{2}/{3}}}d y &= dx\\ \frac {\left (3 c_1 +2 \sqrt {-y}\right ) \left (12 c_1 -24 \sqrt {-y}\right )^{{1}/{3}}}{8}&= x +c_6 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_1 -24 \sqrt {-y}\right )^{{2}/{3}}}{4}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_1^{2}}{4} \end{align*}

Solving Eq. (5)

Integrating gives

\begin{align*} \int \frac {16}{\left (12 c_1 -24 \sqrt {-y}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d y &= dx\\ -\frac {3 \left (1+i \sqrt {3}\right ) \left (12 c_1 -24 \sqrt {-y}\right )^{{1}/{3}} \left (c_1 +\frac {2 \sqrt {-y}}{3}\right )}{16}&= x +c_7 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_1 -24 \sqrt {-y}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}{16}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_1^{2}}{4} \end{align*}

Solving Eq. (6)

Integrating gives

\begin{align*} \int \frac {16}{\left (12 c_1 -24 \sqrt {-y}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d y &= dx\\ \frac {\left (i \sqrt {3}-1\right ) \left (12 c_1 -24 \sqrt {-y}\right )^{{1}/{3}} \left (3 c_1 +2 \sqrt {-y}\right )}{16}&= x +c_8 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_1 -24 \sqrt {-y}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}{16}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_1^{2}}{4} \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} \frac {2 {y^{\prime }}^{{3}/{2}}}{3}+\frac {2 \sqrt {-y y^{\prime }}}{\sqrt {y^{\prime }}} = c_2 \end{align*}

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=\frac {\left (12 c_2 +24 \sqrt {-y}\right )^{{2}/{3}}}{4} \\ \tag{2} y^{\prime }&=\left (-\frac {\left (12 c_2 +24 \sqrt {-y}\right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (12 c_2 +24 \sqrt {-y}\right )^{{1}/{3}}}{4}\right )^{2} \\ \tag{3} y^{\prime }&=\left (-\frac {\left (12 c_2 +24 \sqrt {-y}\right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (12 c_2 +24 \sqrt {-y}\right )^{{1}/{3}}}{4}\right )^{2} \\ \tag{4} y^{\prime }&=\frac {\left (12 c_2 -24 \sqrt {-y}\right )^{{2}/{3}}}{4} \\ \tag{5} y^{\prime }&=\left (-\frac {\left (12 c_2 -24 \sqrt {-y}\right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (12 c_2 -24 \sqrt {-y}\right )^{{1}/{3}}}{4}\right )^{2} \\ \tag{6} y^{\prime }&=\left (-\frac {\left (12 c_2 -24 \sqrt {-y}\right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (12 c_2 -24 \sqrt {-y}\right )^{{1}/{3}}}{4}\right )^{2} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

\begin{align*} \int \frac {4}{\left (12 c_2 +24 \sqrt {-y}\right )^{{2}/{3}}}d y &= dx\\ -\frac {\left (-3 c_2 +2 \sqrt {-y}\right ) \left (12 c_2 +24 \sqrt {-y}\right )^{{1}/{3}}}{8}&= x +c_9 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_2 +24 \sqrt {-y}\right )^{{2}/{3}}}{4}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_2^{2}}{4} \end{align*}

Solving Eq. (2)

Integrating gives

\begin{align*} \int \frac {16}{\left (12 c_2 +24 \sqrt {-y}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d y &= dx\\ -\frac {\left (i \sqrt {3}-1\right ) \left (12 c_2 +24 \sqrt {-y}\right )^{{1}/{3}} \left (-3 c_2 +2 \sqrt {-y}\right )}{16}&= x +\textit {\_C10} \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_2 +24 \sqrt {-y}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}{16}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_2^{2}}{4} \end{align*}

Solving Eq. (3)

Integrating gives

\begin{align*} \int \frac {16}{\left (12 c_2 +24 \sqrt {-y}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d y &= dx\\ -\frac {3 \left (c_2 -\frac {2 \sqrt {-y}}{3}\right ) \left (1+i \sqrt {3}\right ) \left (12 c_2 +24 \sqrt {-y}\right )^{{1}/{3}}}{16}&= x +\textit {\_C11} \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_2 +24 \sqrt {-y}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}{16}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_2^{2}}{4} \end{align*}

Solving Eq. (4)

Integrating gives

\begin{align*} \int \frac {4}{\left (12 c_2 -24 \sqrt {-y}\right )^{{2}/{3}}}d y &= dx\\ \frac {\left (3 c_2 +2 \sqrt {-y}\right ) \left (12 c_2 -24 \sqrt {-y}\right )^{{1}/{3}}}{8}&= x +\textit {\_C12} \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_2 -24 \sqrt {-y}\right )^{{2}/{3}}}{4}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_2^{2}}{4} \end{align*}

Solving Eq. (5)

Integrating gives

\begin{align*} \int \frac {16}{\left (12 c_2 -24 \sqrt {-y}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d y &= dx\\ -\frac {3 \left (1+i \sqrt {3}\right ) \left (12 c_2 -24 \sqrt {-y}\right )^{{1}/{3}} \left (c_2 +\frac {2 \sqrt {-y}}{3}\right )}{16}&= x +\textit {\_C13} \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_2 -24 \sqrt {-y}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}{16}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_2^{2}}{4} \end{align*}

Solving Eq. (6)

Integrating gives

\begin{align*} \int \frac {16}{\left (12 c_2 -24 \sqrt {-y}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d y &= dx\\ \frac {\left (i \sqrt {3}-1\right ) \left (12 c_2 -24 \sqrt {-y}\right )^{{1}/{3}} \left (3 c_2 +2 \sqrt {-y}\right )}{16}&= x +\textit {\_C14} \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\left (12 c_2 -24 \sqrt {-y}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}{16}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_2^{2}}{4} \end{align*}

Will add steps showing solving for IC soon.

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&=-\frac {c_1^{2}}{4}\\ y&=-\frac {c_2^{2}}{4}\\ y&=-\frac {c_1 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_1 \textit {\_Z} +6 c_3 +6 x \right )^{3}}{3}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_1 \textit {\_Z} +6 c_3 +6 x \right )^{2} c_3}{3}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_1 \textit {\_Z} +6 c_3 +6 x \right )^{2} x}{3}-\frac {c_1^{2}}{4}\\ y&=-\frac {c_1 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_1 \textit {\_Z} +6 c_6 +6 x \right )^{3}}{3}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_1 \textit {\_Z} +6 c_6 +6 x \right )^{2} c_6}{3}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_1 \textit {\_Z} +6 c_6 +6 x \right )^{2} x}{3}-\frac {c_1^{2}}{4}\\ y&=-\frac {c_2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_2 \textit {\_Z} +6 \textit {\_C12} +6 x \right )^{3}}{3}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_2 \textit {\_Z} +6 \textit {\_C12} +6 x \right )^{2} \textit {\_C12}}{3}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_2 \textit {\_Z} +6 \textit {\_C12} +6 x \right )^{2} x}{3}-\frac {c_2^{2}}{4}\\ y&=-\frac {c_2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_2 \textit {\_Z} +6 c_9 +6 x \right )^{3}}{3}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_2 \textit {\_Z} +6 c_9 +6 x \right )^{2} c_9}{3}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-6 c_2 \textit {\_Z} +6 c_9 +6 x \right )^{2} x}{3}-\frac {c_2^{2}}{4}\\ y&=-\frac {i \textit {\_C10} \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C10} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C10} -3 x \right )^{2}}{3}-\frac {i x \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C10} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C10} -3 x \right )^{2}}{3}-\frac {c_2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C10} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C10} -3 x \right )^{3}}{3}-\frac {\textit {\_C10} \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C10} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C10} -3 x \right )^{2}}{3}-\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C10} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C10} -3 x \right )^{2}}{3}-\frac {c_2^{2}}{4}\\ y&=-\frac {i \textit {\_C14} \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C14} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C14} -3 x \right )^{2}}{3}-\frac {i x \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C14} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C14} -3 x \right )^{2}}{3}-\frac {c_2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C14} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C14} -3 x \right )^{3}}{3}-\frac {\textit {\_C14} \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C14} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C14} -3 x \right )^{2}}{3}-\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, \textit {\_C14} -3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C14} -3 x \right )^{2}}{3}-\frac {c_2^{2}}{4}\\ y&=-\frac {i c_4 \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_4 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_4 -3 x \right )^{2}}{3}-\frac {i x \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_4 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_4 -3 x \right )^{2}}{3}-\frac {c_1 \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_4 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_4 -3 x \right )^{3}}{3}-\frac {c_4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_4 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_4 -3 x \right )^{2}}{3}-\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_4 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_4 -3 x \right )^{2}}{3}-\frac {c_1^{2}}{4}\\ y&=-\frac {i c_8 \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_8 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_8 -3 x \right )^{2}}{3}-\frac {i x \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_8 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_8 -3 x \right )^{2}}{3}-\frac {c_1 \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_8 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_8 -3 x \right )^{3}}{3}-\frac {c_8 \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_8 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_8 -3 x \right )^{2}}{3}-\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}-3 i \sqrt {3}\, c_8 -3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_8 -3 x \right )^{2}}{3}-\frac {c_1^{2}}{4}\\ y&=\frac {i \textit {\_C11} \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C11} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C11} -3 x \right )^{2}}{3}+\frac {i x \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C11} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C11} -3 x \right )^{2}}{3}-\frac {c_2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C11} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C11} -3 x \right )^{3}}{3}-\frac {\textit {\_C11} \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C11} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C11} -3 x \right )^{2}}{3}-\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C11} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C11} -3 x \right )^{2}}{3}-\frac {c_2^{2}}{4}\\ y&=\frac {i \textit {\_C13} \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C13} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C13} -3 x \right )^{2}}{3}+\frac {i x \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C13} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C13} -3 x \right )^{2}}{3}-\frac {c_2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C13} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C13} -3 x \right )^{3}}{3}-\frac {\textit {\_C13} \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C13} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C13} -3 x \right )^{2}}{3}-\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, \textit {\_C13} +3 i \sqrt {3}\, x -6 c_2 \textit {\_Z} -3 \textit {\_C13} -3 x \right )^{2}}{3}-\frac {c_2^{2}}{4}\\ y&=\frac {i c_5 \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_5 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_5 -3 x \right )^{2}}{3}+\frac {i x \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_5 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_5 -3 x \right )^{2}}{3}-\frac {c_1 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_5 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_5 -3 x \right )^{3}}{3}-\frac {c_5 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_5 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_5 -3 x \right )^{2}}{3}-\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_5 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_5 -3 x \right )^{2}}{3}-\frac {c_1^{2}}{4}\\ y&=\frac {i c_7 \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_7 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_7 -3 x \right )^{2}}{3}+\frac {i x \sqrt {3}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_7 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_7 -3 x \right )^{2}}{3}-\frac {c_1 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_7 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_7 -3 x \right )^{3}}{3}-\frac {c_7 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_7 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_7 -3 x \right )^{2}}{3}-\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3 i \sqrt {3}\, c_7 +3 i \sqrt {3}\, x -6 c_1 \textit {\_Z} -3 c_7 -3 x \right )^{2}}{3}-\frac {c_1^{2}}{4} \end{align*}

1.45.2 Maple step by step solution

1.45.3 Maple trace

Methods for second order ODEs:
 

1.45.4 Maple dsolve solution

Solving time : 0.084 (sec)
Leaf size : 255

dsolve(y(x)*diff(diff(y(x),x),x)^2+diff(y(x),x) = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= c_1 \\ y &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 -3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 +3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \frac {-4 \left (\int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 -3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} \right )-2 i \left (x +c_2 \right ) \sqrt {3}+2 x +2 c_2}{\left (-1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-4 \left (\int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 -3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (x +c_2 \right ) \sqrt {3}+2 x +2 c_2}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-4 \left (\int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 +3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} \right )-2 i \left (x +c_2 \right ) \sqrt {3}+2 x +2 c_2}{\left (-1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-4 \left (\int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 +3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (x +c_2 \right ) \sqrt {3}+2 x +2 c_2}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \end{align*}

1.45.5 Mathematica DSolve solution

Solving time : 60.97 (sec)
Leaf size : 23861

DSolve[{y[x]*D[y[x],{x,2}]^2+D[y[x],x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

Too large to display