2.1.52 Problem 52

Internal problem ID [9122]
Book : Second order enumerated odes
Section : section 1
Problem number : 52
Date solved : Friday, April 25, 2025 at 05:52:17 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved by factoring the differential equation

Time used: 100.239 (sec)

Solve

$$\begin{array}{rl}y{y^{\prime \prime }}^3+y^3{y^{\prime }}^5& =0\end{array}$$

Factoring the ode gives these factors

$$\begin{array}{}\text{(1)}& y& =0\text{(2)}& {y^{\prime }}^5{y}^2+{y^{\prime \prime }}^3& =0\end{array}$$

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for $y$ from

$$\begin{array}{r}y=0\end{array}$$

Solving gives $y=0$

Solving equation (2)

Solved as second order missing x ode

Time used: 100.206 (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

$$\begin{array}{rl}{y}^{\prime }& =p\end{array}$$

Then

$$\begin{array}{rl}{y}^{″}& =\frac{dp}{dx}\\ & =\frac{dp}{dy}\frac{dy}{dx}\\ & =p\frac{dp}{dy}\end{array}$$

Hence the ode becomes

$$\begin{array}{r}p{\left(y\right)}^5{y}^2+p{\left(y\right)}^3{\left(\frac{d}{dy}p\left(y\right)\right)}^3=0\end{array}$$

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

Solved by factoring the differential equation

Time used: 0.938 (sec)

Solve

$$\begin{array}{rl}{p}^5{y}^2+p^3{p^{\prime }}^3& =0\end{array}$$

Factoring the ode gives these factors

$$\begin{array}{}\text{(1)}& p^3& =0\text{(2)}& p^2{y}^2+{p^{\prime }}^3& =0\end{array}$$

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for $p$ from

$$\begin{array}{r}{p}^3=0\end{array}$$

Solving gives $p=0$

Solving equation (2)

The ode has the form

$$\begin{array}{}\text{(1)}& (p^{\prime }{)}^{\frac{n}{m}}& =f(y)g(p)\end{array}$$

Where $n=3,m=1,f=-y^2,g=p^2$. Hence the ode is

$$\begin{array}{rl}(p^{\prime }{)}^3& =-p^2{y}^2\end{array}$$

Solving for $p^{\prime }$ from (1) gives

$$\begin{array}{rl}{p}^{\prime }& ={\left(fg\right)}^{1/3}\\ p^{\prime }& =-\frac{{\left(fg\right)}^{1/3}}{2}+\frac{i\sqrt{3}{\left(fg\right)}^{1/3}}{2}\\ p^{\prime }& =-\frac{{\left(fg\right)}^{1/3}}{2}-\frac{i\sqrt{3}{\left(fg\right)}^{1/3}}{2}\end{array}$$

To be able to solve as separable ode, we have to now assume that $f>0,g>0$.

$$\begin{array}{rl}-y^2& >0\\ p^2& >0\end{array}$$

Under the above assumption the differential equations become separable and can be written as

$$\begin{array}{rl}{p}^{\prime }& =f^{1/3}{g}^{1/3}\\ p^{\prime }& =\frac{f^{1/3}{g}^{1/3}(-1+i\sqrt{3})}{2}\\ p^{\prime }& =-\frac{f^{1/3}{g}^{1/3}(1+i\sqrt{3})}{2}\end{array}$$

Therefore

$$\begin{array}{rl}\frac{1}{g^{1/3}}dp& =\left(f^{1/3}\right)dy\\ \frac{2}{g^{1/3}(-1+i\sqrt{3})}dp& =\left(f^{1/3}\right)dy\\ -\frac{2}{g^{1/3}(1+i\sqrt{3})}dp& =\left(f^{1/3}\right)dy\end{array}$$

Replacing $f(y),g(p)$ by their values gives

$$\begin{array}{rl}\frac{1}{{\left(p^2\right)}^{1/3}}dp& =\left({(-y^2)}^{1/3}\right)dy\\ \frac{2}{{\left(p^2\right)}^{1/3}(-1+i\sqrt{3})}dp& =\left({(-y^2)}^{1/3}\right)dy\\ -\frac{2}{{\left(p^2\right)}^{1/3}(1+i\sqrt{3})}dp& =\left({(-y^2)}^{1/3}\right)dy\end{array}$$

Integrating now gives the following solutions

$$\begin{array}{rl}\int \frac{1}{{\left(p^2\right)}^{1/3}}dp& =\int {(-y^2)}^{1/3}dy+c_1\\ \frac{3{\left(p^2\right)}^{2/3}}{p}& =\frac{3y{(-y^2)}^{1/3}}{5}\\ \int \frac{2}{{\left(p^2\right)}^{1/3}(-1+i\sqrt{3})}dp& =\int {(-y^2)}^{1/3}dy+c_1\\ -\frac{3{\left(p^2\right)}^{2/3}(1+i\sqrt{3})}{2p}& =\frac{3y{(-y^2)}^{1/3}}{5}\\ \int -\frac{2}{{\left(p^2\right)}^{1/3}(1+i\sqrt{3})}dp& =\int {(-y^2)}^{1/3}dy+c_1\\ \frac{3{\left(p^2\right)}^{2/3}(-1+i\sqrt{3})}{2p}& =\frac{3y{(-y^2)}^{1/3}}{5}\end{array}$$

Therefore

$$\begin{array}{rl}\frac{3{\left(p^2\right)}^{2/3}}{p}& =\frac{3y{(-y^2)}^{1/3}}{5}+c_1\\ p& =-\frac{y^5}{125}+\frac{{(-y^2)}^{2/3}{c}_1{y}^2}{25}+\frac{{(-y^2)}^{1/3}{c}_1^{2}y}{15}+\frac{c_1^3}{27}\\ p& =-\frac{y^5}{125}+\frac{{(-y^2)}^{2/3}{c}_1{y}^2}{25}+\frac{{(-y^2)}^{1/3}{c}_1^{2}y}{15}+\frac{c_1^3}{27}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}\frac{3{\left(p^2\right)}^{2/3}}{p}& =\frac{3y{(-y^2)}^{1/3}}{5}+c_1\\ p& =0\\ p& =-\frac{y^5}{125}+\frac{{(-y^2)}^{2/3}{c}_1{y}^2}{25}+\frac{{(-y^2)}^{1/3}{c}_1^{2}y}{15}+\frac{c_1^3}{27}\end{array}$$

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

$$\begin{array}{r}\frac{3{\left({y^{\prime }}^2\right)}^{2/3}}{y^{\prime }}=\frac{3y{(-y^2)}^{1/3}}{5}+c_1\end{array}$$

Let $p=y^{\prime }$ the ode becomes

$$\begin{array}{r}\frac{3{\left(p^2\right)}^{2/3}}{p}=\frac{3y{(-y^2)}^{1/3}}{5}+c_1\end{array}$$

Solving for $y$ from the above results in

$$\begin{array}{}\text{(1)}& y& =\frac{5^{3/5}{3}^{2/5}{\left((27{\left(p^2\right)}^{4/3}{c}_1-9{\left(p^2\right)}^{2/3}{c}_1^{2}p+c_1^3{p}^2-27p^3)p^3\right)}^{1/5}}{3p}\text{(2)}& y& =\frac{(\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5+\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{\left((27{\left(p^2\right)}^{4/3}{c}_1-9{\left(p^2\right)}^{2/3}{c}_1^{2}p+c_1^3{p}^2-27p^3)p^3\right)}^{1/5}}{3p}\text{(3)}& y& =\frac{(-\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{\left((27{\left(p^2\right)}^{4/3}{c}_1-9{\left(p^2\right)}^{2/3}{c}_1^{2}p+c_1^3{p}^2-27p^3)p^3\right)}^{1/5}}{3p}\text{(4)}& y& =\frac{(-\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{\left((27{\left(p^2\right)}^{4/3}{c}_1-9{\left(p^2\right)}^{2/3}{c}_1^{2}p+c_1^3{p}^2-27p^3)p^3\right)}^{1/5}}{3p}\text{(5)}& y& =\frac{(\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5+\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{\left((27{\left(p^2\right)}^{4/3}{c}_1-9{\left(p^2\right)}^{2/3}{c}_1^{2}p+c_1^3{p}^2-27p^3)p^3\right)}^{1/5}}{3p}\end{array}$$

This has the form

$$\begin{array}{}\text{(*)}& y=xf(p)+g(p)\end{array}$$

Where $f,g$ are functions of $p=y^{\prime }(x)$. Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =\frac{5^{3/5}{3}^{2/5}{(-27(\frac{c_1^2{\left(p^2\right)}^{2/3}}{3}+p(-\frac{c_1^3}{27}-c_1{\left(p^2\right)}^{1/3}+p))p^4)}^{1/5}}{3p}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p=(-\frac{45^{3/5}{3}^{2/5}{p}^4{c}_1^2}{5{(-9p^4{c}_1^2{\left(p^2\right)}^{2/3}+p^5{c}_1^3+27p^5{c}_1{\left(p^2\right)}^{1/3}-27p^6)}^{4/5}{\left(p^2\right)}^{1/3}}+\frac{5^{3/5}{3}^{2/5}{p}^3{c}_1^3}{3{(-9p^4{c}_1^2{\left(p^2\right)}^{2/3}+p^5{c}_1^3+27p^5{c}_1{\left(p^2\right)}^{1/3}-27p^6)}^{4/5}}+\frac{95^{3/5}{3}^{2/5}{p}^3{c}_1{\left(p^2\right)}^{1/3}}{{(-9p^4{c}_1^2{\left(p^2\right)}^{2/3}+p^5{c}_1^3+27p^5{c}_1{\left(p^2\right)}^{1/3}-27p^6)}^{4/5}}-\frac{545^{3/5}{3}^{2/5}{p}^4}{5{(-9p^4{c}_1^2{\left(p^2\right)}^{2/3}+p^5{c}_1^3+27p^5{c}_1{\left(p^2\right)}^{1/3}-27p^6)}^{4/5}}+\frac{65^{3/5}{3}^{2/5}{p}^5{c}_1}{5{(-9p^4{c}_1^2{\left(p^2\right)}^{2/3}+p^5{c}_1^3+27p^5{c}_1{\left(p^2\right)}^{1/3}-27p^6)}^{4/5}{\left(p^2\right)}^{2/3}}-\frac{125^{3/5}{3}^{2/5}{p}^2{c}_1^2{\left(p^2\right)}^{2/3}}{5{(-9p^4{c}_1^2{\left(p^2\right)}^{2/3}+p^5{c}_1^3+27p^5{c}_1{\left(p^2\right)}^{1/3}-27p^6)}^{4/5}}-\frac{5^{3/5}{3}^{2/5}{(-9p^4{c}_1^2{\left(p^2\right)}^{2/3}+p^5{c}_1^3+27p^5{c}_1{\left(p^2\right)}^{1/3}-27p^6)}^{1/5}}{3p^2})p^{\prime }\left(x\right)\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=-\frac{{(-c_1^3)}^{1/5}{5}^{3/5}{3}^{2/5}}{3}\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& p^{\prime }\left(x\right)=\frac{p\left(x\right)}{-\frac{45^{3/5}{3}^{2/5}p{\left(x\right)}^4{c}_1^2}{5{(-9p{\left(x\right)}^4{c}_1^2{\left(p{\left(x\right)}^2\right)}^{2/3}+p{\left(x\right)}^5{c}_1^3+27p{\left(x\right)}^5{c}_1{\left(p{\left(x\right)}^2\right)}^{1/3}-27p{\left(x\right)}^6)}^{4/5}{\left(p{\left(x\right)}^2\right)}^{1/3}}+\frac{5^{3/5}{3}^{2/5}p{\left(x\right)}^3{c}_1^3}{3{(-9p{\left(x\right)}^4{c}_1^2{\left(p{\left(x\right)}^2\right)}^{2/3}+p{\left(x\right)}^5{c}_1^3+27p{\left(x\right)}^5{c}_1{\left(p{\left(x\right)}^2\right)}^{1/3}-27p{\left(x\right)}^6)}^{4/5}}+\frac{95^{3/5}{3}^{2/5}p{\left(x\right)}^3{c}_1{\left(p{\left(x\right)}^2\right)}^{1/3}}{{(-9p{\left(x\right)}^4{c}_1^2{\left(p{\left(x\right)}^2\right)}^{2/3}+p{\left(x\right)}^5{c}_1^3+27p{\left(x\right)}^5{c}_1{\left(p{\left(x\right)}^2\right)}^{1/3}-27p{\left(x\right)}^6)}^{4/5}}-\frac{545^{3/5}{3}^{2/5}p{\left(x\right)}^4}{5{(-9p{\left(x\right)}^4{c}_1^2{\left(p{\left(x\right)}^2\right)}^{2/3}+p{\left(x\right)}^5{c}_1^3+27p{\left(x\right)}^5{c}_1{\left(p{\left(x\right)}^2\right)}^{1/3}-27p{\left(x\right)}^6)}^{4/5}}+\frac{65^{3/5}{3}^{2/5}p{\left(x\right)}^5{c}_1}{5{(-9p{\left(x\right)}^4{c}_1^2{\left(p{\left(x\right)}^2\right)}^{2/3}+p{\left(x\right)}^5{c}_1^3+27p{\left(x\right)}^5{c}_1{\left(p{\left(x\right)}^2\right)}^{1/3}-27p{\left(x\right)}^6)}^{4/5}{\left(p{\left(x\right)}^2\right)}^{2/3}}-\frac{125^{3/5}{3}^{2/5}p{\left(x\right)}^2{c}_1^2{\left(p{\left(x\right)}^2\right)}^{2/3}}{5{(-9p{\left(x\right)}^4{c}_1^2{\left(p{\left(x\right)}^2\right)}^{2/3}+p{\left(x\right)}^5{c}_1^3+27p{\left(x\right)}^5{c}_1{\left(p{\left(x\right)}^2\right)}^{1/3}-27p{\left(x\right)}^6)}^{4/5}}-\frac{5^{3/5}{3}^{2/5}{(-9p{\left(x\right)}^4{c}_1^2{\left(p{\left(x\right)}^2\right)}^{2/3}+p{\left(x\right)}^5{c}_1^3+27p{\left(x\right)}^5{c}_1{\left(p{\left(x\right)}^2\right)}^{1/3}-27p{\left(x\right)}^6)}^{1/5}}{3p{\left(x\right)}^2}}\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^{p\left(x\right)}\frac{5^{3/5}{3}^{2/5}{\tau }^3(-c_1^2{\left({\tau }^2\right)}^{1/3}+6c_1\tau -9{\left({\tau }^2\right)}^{2/3})}{5{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+\tau (c_1^3+27c_1{\left({\tau }^2\right)}^{1/3}-27\tau )){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau =x+c_2$$

Substituing the above solution for $p$ in (2A) gives

$$\begin{array}{rl}y& =\frac{5^{3/5}{3}^{2/5}{(-27(\frac{c_1^2{\left(\mathrm{RootOf}{(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau ){675}^{4/5}}{225{\left({\tau }^2\right)}^{2/3}{\left({(-9c_1^2{\left({\tau }^2\right)}^{2/3}+c_1^3\tau +27{\left({\tau }^2\right)}^{1/3}{c}_1\tau -27{\tau }^2)}^4{\tau }^{16}\right)}^{1/5}}d\tau +x+c_2)}^2\right)}^{2/3}}{3}+\mathrm{RootOf}(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{5^{3/5}{3}^{2/5}{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau )}{5{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+c_1^3\tau +27{\left({\tau }^2\right)}^{1/3}{c}_1\tau -27{\tau }^2){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau +x+c_2)(-\frac{c_1^3}{27}-c_1{\left(\mathrm{RootOf}{(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau ){675}^{4/5}}{225{\left({\tau }^2\right)}^{2/3}{\left({(-9c_1^2{\left({\tau }^2\right)}^{2/3}+c_1^3\tau +27{\left({\tau }^2\right)}^{1/3}{c}_1\tau -27{\tau }^2)}^4{\tau }^{16}\right)}^{1/5}}d\tau +x+c_2)}^2\right)}^{1/3}+\mathrm{RootOf}(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{5^{3/5}{3}^{2/5}{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau )}{5{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+c_1^3\tau +27{\left({\tau }^2\right)}^{1/3}{c}_1\tau -27{\tau }^2){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau +x+c_2)))\mathrm{RootOf}{(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{5^{3/5}{3}^{2/5}{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau )}{5{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+c_1^3\tau +27{\left({\tau }^2\right)}^{1/3}{c}_1\tau -27{\tau }^2){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau +x+c_2)}^4)}^{1/5}}{3\mathrm{RootOf}(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{5^{3/5}{3}^{2/5}{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau )}{5{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+c_1^3\tau +27{\left({\tau }^2\right)}^{1/3}{c}_1\tau -27{\tau }^2){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau +x+c_2)}\end{array}$$

Solving ode 2A

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =\frac{(i\sqrt{2}\sqrt{5+\sqrt{5}}+\sqrt{5}-1)5^{3/5}{3}^{2/5}{(-27(\frac{c_1^2{\left(p^2\right)}^{2/3}}{3}+p(-\frac{c_1^3}{27}-c_1{\left(p^2\right)}^{1/3}+p))p^4)}^{1/5}}{12p}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& \text{Expression too large to display}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=-\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}-5^{3/5}+55^{1/10})}{12}\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& \text{Expression too large to display}\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^{p\left(x\right)}-\frac{3^{2/5}{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau )(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}-5^{3/5}+55^{1/10})}{20{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+\tau (c_1^3+27c_1{\left({\tau }^2\right)}^{1/3}-27\tau )){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau =x+c_3$$

Substituing the above solution for $p$ in (2A) gives

$$\begin{array}{r}\text{Expression too large to display}\end{array}$$

Solving ode 3A

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =-\frac{(-i\sqrt{2}\sqrt{5-\sqrt{5}}+\sqrt{5}+1)5^{3/5}{3}^{2/5}{(-27(\frac{c_1^2{\left(p^2\right)}^{2/3}}{3}+p(-\frac{c_1^3}{27}-c_1{\left(p^2\right)}^{1/3}+p))p^4)}^{1/5}}{12p}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& \text{Expression too large to display}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=-\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}-5^{3/5}-55^{1/10})}{12}\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& \text{Expression too large to display}\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^{p\left(x\right)}\frac{3^{2/5}{\tau }^3(-c_1^2{\left({\tau }^2\right)}^{1/3}+6c_1\tau -9{\left({\tau }^2\right)}^{2/3})(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}-5^{3/5}-55^{1/10})}{20{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+\tau (c_1^3+27c_1{\left({\tau }^2\right)}^{1/3}-27\tau )){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau =x+c_4$$

Substituing the above solution for $p$ in (2A) gives

$$\begin{array}{r}\text{Expression too large to display}\end{array}$$

Solving ode 4A

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =-\frac{(i\sqrt{2}\sqrt{5-\sqrt{5}}+\sqrt{5}+1)5^{3/5}{3}^{2/5}{(-27(\frac{c_1^2{\left(p^2\right)}^{2/3}}{3}+p(-\frac{c_1^3}{27}-c_1{\left(p^2\right)}^{1/3}+p))p^4)}^{1/5}}{12p}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& \text{Expression too large to display}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}+5^{3/5}+55^{1/10})}{12}\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& \text{Expression too large to display}\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^{p\left(x\right)}\frac{3^{2/5}{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau )(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}+5^{3/5}+55^{1/10})}{20{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+\tau (c_1^3+27c_1{\left({\tau }^2\right)}^{1/3}-27\tau )){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau =x+c_5$$

Substituing the above solution for $p$ in (2A) gives

$$\begin{array}{r}\text{Expression too large to display}\end{array}$$

Solving ode 5A

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =-\frac{(i\sqrt{2}\sqrt{5+\sqrt{5}}-\sqrt{5}+1)5^{3/5}{3}^{2/5}{(-27(\frac{c_1^2{\left(p^2\right)}^{2/3}}{3}+p(-\frac{c_1^3}{27}-c_1{\left(p^2\right)}^{1/3}+p))p^4)}^{1/5}}{12p}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& \text{Expression too large to display}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}+5^{3/5}-55^{1/10})}{12}\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& \text{Expression too large to display}\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^{p\left(x\right)}\frac{3^{2/5}{\tau }^3(c_1^2{\left({\tau }^2\right)}^{1/3}+9{\left({\tau }^2\right)}^{2/3}-6c_1\tau )(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}+5^{3/5}-55^{1/10})}{20{\left((-9c_1^2{\left({\tau }^2\right)}^{2/3}+\tau (c_1^3+27c_1{\left({\tau }^2\right)}^{1/3}-27\tau )){\tau }^4\right)}^{4/5}{\left({\tau }^2\right)}^{2/3}}d\tau =x+c_6$$

Substituing the above solution for $p$ in (2A) gives

$$\begin{array}{r}\text{Expression too large to display}\end{array}$$

Which simplifies to

$$\begin{array}{rl}y& =-\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}-5^{3/5}-55^{1/10})}{12}\\ y& =\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}+5^{3/5}+55^{1/10})}{12}\\ y& =-\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}-5^{3/5}+55^{1/10})}{12}\\ y& =\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}+5^{3/5}-55^{1/10})}{12}\\ y& =-\frac{{(-c_1^3)}^{1/5}{5}^{3/5}{3}^{2/5}}{3}\\ \text{Expression too large to display}\\ \text{Expression too large to display}\\ \text{Expression too large to display}\\ \text{Expression too large to display}\\ \text{Expression too large to display}\end{array}$$

The solution

$$\text{Expression too large to display}$$

was found not to satisfy the ode or the IC. Hence it is removed. The solution

$$\text{Expression too large to display}$$

was found not to satisfy the ode or the IC. Hence it is removed. The solution

$$\text{Expression too large to display}$$

was found not to satisfy the ode or the IC. Hence it is removed. The solution

$$\text{Expression too large to display}$$

was found not to satisfy the ode or the IC. Hence it is removed. The solution

$$\text{Expression too large to display}$$

was found not to satisfy the ode or the IC. Hence it is removed. For solution (2) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=0\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int 0dx+c_7\\ y& =c_7\end{array}$$

For solution (3) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=-\frac{y^5}{125}+\frac{y^2{(-y^2)}^{2/3}{c}_1}{25}+\frac{y{(-y^2)}^{1/3}{c}_1^2}{15}+\frac{c_1^3}{27}\end{array}$$

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^y\frac{1}{-\frac{{\tau }^5}{125}+\frac{{(-{\tau }^2)}^{2/3}{c}_1{\tau }^2}{25}+\frac{{(-{\tau }^2)}^{1/3}{c}_1^2\tau }{15}+\frac{c_1^3}{27}}d\tau =x+c_8$$

Singular solutions are found by solving

$$\begin{array}{rl}-\frac{y^5}{125}+\frac{{(-y^2)}^{2/3}{c}_1{y}^2}{25}+\frac{{(-y^2)}^{1/3}{c}_1^{2}y}{15}+\frac{c_1^3}{27}& =0\end{array}$$

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.

$$\begin{array}{r}y=\frac{(-\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{c}_1^{3/5}}{3}\\ y=\frac{(-\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{c}_1^{3/5}}{3}\end{array}$$

Which simplifies to

$$\begin{array}{rl}3375{\int }_{}^y\frac{1}{135{(-{\tau }^2)}^{2/3}{c}_1{\tau }^2-27{\tau }^5+225{(-{\tau }^2)}^{1/3}{c}_1^2\tau +125c_1^3}d\tau & =x+c_8\\ y& =-\frac{(i\sqrt{2}\sqrt{5-\sqrt{5}}+\sqrt{5}+1)5^{3/5}{3}^{2/5}{c}_1^{3/5}}{12}\\ y& =\frac{(-\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{c}_1^{3/5}}{3}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}3375{\int }_{}^y\frac{1}{135{(-{\tau }^2)}^{2/3}{c}_1{\tau }^2-27{\tau }^5+225{(-{\tau }^2)}^{1/3}{c}_1^2\tau +125c_1^3}d\tau & =x+c_8\\ y& =c_7\\ y& =-\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}-5^{3/5}-55^{1/10})}{12}\\ y& =\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}+5^{3/5}+55^{1/10})}{12}\\ y& =-\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}-5^{3/5}+55^{1/10})}{12}\\ y& =\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}+5^{3/5}-55^{1/10})}{12}\\ y& =-\frac{{(-c_1^3)}^{1/5}{5}^{3/5}{3}^{2/5}}{3}\\ y& =\frac{(-\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{c}_1^{3/5}}{3}\\ y& =-\frac{(i\sqrt{2}\sqrt{5-\sqrt{5}}+\sqrt{5}+1)5^{3/5}{3}^{2/5}{c}_1^{3/5}}{12}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}3375{\int }_{}^y\frac{1}{135{(-{\tau }^2)}^{2/3}{c}_1{\tau }^2-27{\tau }^5+225{(-{\tau }^2)}^{1/3}{c}_1^2\tau +125c_1^3}d\tau & =x+c_8\\ y& =0\\ y& =c_7\\ y& =-\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}-5^{3/5}-55^{1/10})}{12}\\ y& =\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5-\sqrt{5}}+5^{3/5}+55^{1/10})}{12}\\ y& =-\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}-5^{3/5}+55^{1/10})}{12}\\ y& =\frac{{(-c_1^3)}^{1/5}{3}^{2/5}(i{5}^{3/5}\sqrt{2}\sqrt{5+\sqrt{5}}+5^{3/5}-55^{1/10})}{12}\\ y& =-\frac{{(-c_1^3)}^{1/5}{5}^{3/5}{3}^{2/5}}{3}\\ y& =\frac{(-\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})5^{3/5}{3}^{2/5}{c}_1^{3/5}}{3}\\ y& =-\frac{(i\sqrt{2}\sqrt{5-\sqrt{5}}+\sqrt{5}+1)5^{3/5}{3}^{2/5}{c}_1^{3/5}}{12}\end{array}$$

✓ Maple. Time used: 0.275 (sec). Leaf size: 208

ode:=y(x)*diff(diff(y(x),x),x)^3+y(x)^3*diff(y(x),x)^5 = 0; 
dsolve(ode,y(x), singsol=all);
 

$$\begin{array}{rl}y& =0\\ y& =c_1\\ {\int }_{}^y\frac{1}{\mathrm{RootOf}(5{\int }_{\mathit{\text{\_g}}}^{\mathit{\text{\_Z}}}\frac{1}{\mathit{\text{\_a}}{(-{\mathit{\text{\_a}}}^2{\mathit{\text{\_f}}}^2)}^{1/3}-5\mathit{\text{\_f}}}d\mathit{\text{\_f}}-\ln ({\mathit{\text{\_a}}}^5+125)+5c_1)}d\mathit{\text{\_a}}-x-c_2& =0\\ {\int }_{}^y\frac{1}{\mathrm{RootOf}(-i\ln ({\mathit{\text{\_a}}}^5+125)+\ln ({\mathit{\text{\_a}}}^5+125)\sqrt{3}+20{\int }_{\mathit{\text{\_g}}}^{\mathit{\text{\_Z}}}\frac{1}{2i\mathit{\text{\_a}}{(-{\mathit{\text{\_a}}}^2{\mathit{\text{\_f}}}^2)}^{1/3}+5i\mathit{\text{\_f}}+5\sqrt{3}\mathit{\text{\_f}}}d\mathit{\text{\_f}}-20c_1)}d\mathit{\text{\_a}}-x-c_2& =0\\ {\int }_{}^y\frac{1}{\mathrm{RootOf}(\ln ({\mathit{\text{\_a}}}^5+125)\sqrt{3}+i\ln ({\mathit{\text{\_a}}}^5+125)+20{\int }_{\mathit{\text{\_g}}}^{\mathit{\text{\_Z}}}\frac{1}{-2i\mathit{\text{\_a}}{(-{\mathit{\text{\_a}}}^2{\mathit{\text{\_f}}}^2)}^{1/3}+5\sqrt{3}\mathit{\text{\_f}}-5i\mathit{\text{\_f}}}d\mathit{\text{\_f}}-20c_1)}d\mathit{\text{\_a}}-x-c_2& =0\end{array}$$

Maple trace

Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 3 solutions were found. Trying to solve e\ 
ach resulting ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      -> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)-(-_a^2*_b(_a)^2)^ 
(1/3)*_b(_a) = 0, _b(_a), HINT = [[_a, 5*_b]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[_a, 5*_b] 
         1st order, trying the canonical coordinates of the invariance group 
            -> Calling odsolve with the ODE, diff(y(x),x) = 5*y(x)/x, y(x) 
               *** Sublevel 5 *** 
               Methods for first order ODEs: 
               --- Trying classification methods --- 
               trying a quadrature 
               trying 1st order linear 
               <- 1st order linear successful 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      -> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+1/2*(-_a^2*_b(_a) 
^2)^(1/3)*(1+I*3^(1/2))*_b(_a) = 0, _b(_a), HINT = [[_a, 5*_b]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[_a, 5*_b] 
         1st order, trying the canonical coordinates of the invariance group 
            -> Calling odsolve with the ODE, diff(y(x),x) = 5*y(x)/x, y(x) 
               *** Sublevel 5 *** 
               Methods for first order ODEs: 
               --- Trying classification methods --- 
               trying a quadrature 
               trying 1st order linear 
               <- 1st order linear successful 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      -> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)-1/2*(-_a^2*_b(_a) 
^2)^(1/3)*(I*3^(1/2)-1)*_b(_a) = 0, _b(_a), HINT = [[_a, 5*_b]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[_a, 5*_b] 
         1st order, trying the canonical coordinates of the invariance group 
            -> Calling odsolve with the ODE, diff(y(x),x) = 5*y(x)/x, y(x) 
               *** Sublevel 5 *** 
               Methods for first order ODEs: 
               --- Trying classification methods --- 
               trying a quadrature 
               trying 1st order linear 
               <- 1st order linear successful 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
-> Calling odsolve with the ODE, diff(y(x),x) = 0, y(x), singsol = none 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful
 

✓ Mathematica. Time used: 24.151 (sec). Leaf size: 449

ode=y[x]*D[y[x],{x,2}]^3+y[x]^3*D[y[x],x]^5==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}\text{Solution too large to show}\end{array}$$

✓ Sympy. Time used: 13.171 (sec). Leaf size: 3

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)**3*Derivative(y(x), x)**5 + y(x)*Derivative(y(x), (x, 2))**3,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=0$$