Internal problem ID [4314]
Internal file name [OUTPUT/3807_Sunday_June_05_2022_11_05_12_AM_83543797/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 36
Problem number: 1102.
ODE order: 1.
ODE degree: 6.
The type(s) of ODE detected by this program : "first_order_nonlinear_p_but_separable"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
\[ \boxed {{y^{\prime }}^{6}+f \left (x \right ) \left (y-a \right )^{5} \left (y-b \right )^{4}=0} \]
The ode has the form \begin {align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end {align*}
Where \(n=6, m=1, f=f \left (x \right ) , g=\left (b -y \right )^{4} \left (a -y \right )^{5}\). Hence the ode is \begin {align*} (y')^{6} &= f \left (x \right ) \left (b -y \right )^{4} \left (a -y \right )^{5} \end {align*}
Solving for \(y^{\prime }\) from (1) gives \begin {align*} y^{\prime } &=\left (f g \right )^{\frac {1}{6}}\\ y^{\prime } &=\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (f g \right )^{\frac {1}{6}}\\ y^{\prime } &=\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (f g \right )^{\frac {1}{6}}\\ y^{\prime } &=-\left (f g \right )^{\frac {1}{6}}\\ y^{\prime } &=\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (f g \right )^{\frac {1}{6}}\\ y^{\prime } &=\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (f g \right )^{\frac {1}{6}} \end {align*}
To be able to solve as separable ode, we have to now assume that \(f>0,g>0\). \begin {align*} f \left (x \right ) &> 0\\ \left (b -y \right )^{4} \left (a -y \right )^{5} &> 0 \end {align*}
Under the above assumption the differential equations become separable and can be written as \begin {align*} y^{\prime } &=f^{\frac {1}{6}} g^{\frac {1}{6}}\\ y^{\prime } &=\frac {\left (1+i \sqrt {3}\right ) f^{\frac {1}{6}} g^{\frac {1}{6}}}{2}\\ y^{\prime } &=\frac {\left (i \sqrt {3}-1\right ) f^{\frac {1}{6}} g^{\frac {1}{6}}}{2}\\ y^{\prime } &=-f^{\frac {1}{6}} g^{\frac {1}{6}}\\ y^{\prime } &=-\frac {\left (1+i \sqrt {3}\right ) f^{\frac {1}{6}} g^{\frac {1}{6}}}{2}\\ y^{\prime } &=-\frac {\left (i \sqrt {3}-1\right ) f^{\frac {1}{6}} g^{\frac {1}{6}}}{2} \end {align*}
Therefore \begin {align*} \frac {1}{g^{\frac {1}{6}}} \, dy &= \left (f^{\frac {1}{6}}\right )\,dx\\ \frac {2}{\left (1+i \sqrt {3}\right ) g^{\frac {1}{6}}} \, dy &= \left (f^{\frac {1}{6}}\right )\,dx\\ \frac {2}{\left (i \sqrt {3}-1\right ) g^{\frac {1}{6}}} \, dy &= \left (f^{\frac {1}{6}}\right )\,dx\\ -\frac {1}{g^{\frac {1}{6}}} \, dy &= \left (f^{\frac {1}{6}}\right )\,dx\\ -\frac {2}{\left (1+i \sqrt {3}\right ) g^{\frac {1}{6}}} \, dy &= \left (f^{\frac {1}{6}}\right )\,dx\\ -\frac {2}{\left (i \sqrt {3}-1\right ) g^{\frac {1}{6}}} \, dy &= \left (f^{\frac {1}{6}}\right )\,dx \end {align*}
Replacing \(f(x),g(y)\) by their values gives \begin {align*} \frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}} \, dy &= \left (f \left (x \right )^{\frac {1}{6}}\right )\,dx\\ \frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}} \, dy &= \left (f \left (x \right )^{\frac {1}{6}}\right )\,dx\\ \frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}} \, dy &= \left (f \left (x \right )^{\frac {1}{6}}\right )\,dx\\ -\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}} \, dy &= \left (f \left (x \right )^{\frac {1}{6}}\right )\,dx\\ -\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}} \, dy &= \left (f \left (x \right )^{\frac {1}{6}}\right )\,dx\\ -\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}} \, dy &= \left (f \left (x \right )^{\frac {1}{6}}\right )\,dx \end {align*}
Integrating now gives the solutions. \begin {align*} \int \frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d y &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int \frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d y &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int \frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d y &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int -\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d y &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int -\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d y &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int -\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d y &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \end {align*}
Integrating gives \begin {align*} \int _{}^{y}\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int _{}^{y}\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int _{}^{y}\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int _{}^{y}-\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int _{}^{y}-\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}\\ \int _{}^{y}-\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \end {align*}
Therefore \begin{align*} \int _{}^{y}\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \int _{}^{y}\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \int _{}^{y}\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \int _{}^{y}-\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \int _{}^{y}-\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \int _{}^{y}-\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \end{align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \tag{2} \int _{}^{y}\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \tag{3} \int _{}^{y}\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \tag{4} \int _{}^{y}-\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \tag{5} \int _{}^{y}-\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \tag{6} \int _{}^{y}-\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{6}}d x +c_{1} \\ \end{align*}
Verification of solutions
\[
\int _{}^{y}\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}
\] Verified OK. {0 < (b-y)^4*(a-y)^5, 0 < f(x)}
\[
\int _{}^{y}\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}
\] Verified OK. {0 < (b-y)^4*(a-y)^5, 0 < f(x)}
\[
\int _{}^{y}\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}
\] Verified OK. {0 < (b-y)^4*(a-y)^5, 0 < f(x)}
\[
\int _{}^{y}-\frac {1}{\left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}
\] Verified OK. {0 < (b-y)^4*(a-y)^5, 0 < f(x)}
\[
\int _{}^{y}-\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}
\] Verified OK. {0 < (b-y)^4*(a-y)^5, 0 < f(x)}
\[
\int _{}^{y}-\frac {2}{\left (i \sqrt {3}-1\right ) \left (\left (b -y \right )^{4} \left (a -y \right )^{5}\right )^{\frac {1}{6}}}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{6}}d x +c_{1}
\] Verified OK. {0 < (b-y)^4*(a-y)^5, 0 < f(x)}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{6}+f \left (x \right ) \left (y-a \right )^{5} \left (y-b \right )^{4}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\mathit {RootOf}\left (-5 f \left (x \right ) y^{8} a +10 f \left (x \right ) y^{7} a^{2}-4 f \left (x \right ) y^{8} b -10 f \left (x \right ) y^{6} a^{3}+6 f \left (x \right ) y^{7} b^{2}-4 f \left (x \right ) y^{6} b^{3}+5 f \left (x \right ) y^{5} a^{4}+f \left (x \right ) y^{5} b^{4}-f \left (x \right ) y^{4} a^{5}-f \left (x \right ) a^{5} b^{4}+20 f \left (x \right ) y^{7} a b -40 f \left (x \right ) y^{6} a^{2} b -30 f \left (x \right ) y^{6} a \,b^{2}+40 f \left (x \right ) y^{5} a^{3} b +60 f \left (x \right ) y^{5} a^{2} b^{2}+20 f \left (x \right ) y^{5} a \,b^{3}-20 f \left (x \right ) y^{4} a^{4} b -60 f \left (x \right ) y^{4} a^{3} b^{2}-40 f \left (x \right ) y^{4} a^{2} b^{3}-5 f \left (x \right ) y^{4} a \,b^{4}+4 f \left (x \right ) y^{3} a^{5} b +30 f \left (x \right ) y^{3} a^{4} b^{2}+40 f \left (x \right ) y^{3} a^{3} b^{3}+10 f \left (x \right ) y^{3} a^{2} b^{4}-6 f \left (x \right ) y^{2} a^{5} b^{2}-20 f \left (x \right ) y^{2} a^{4} b^{3}-10 f \left (x \right ) y^{2} a^{3} b^{4}+4 f \left (x \right ) y a^{5} b^{3}+5 f \left (x \right ) y a^{4} b^{4}+f \left (x \right ) y^{9}+\textit {\_Z}^{6}\right ) \end {array} \]
Maple trace
`Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying simple symmetries for implicit equations --- Trying classification methods --- trying homogeneous types: trying exact <- exact successful`
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 69
dsolve(diff(y(x),x)^6+f(x)*(y(x)-a)^5*(y(x)-b)^4 = 0,y(x), singsol=all)
\[ \int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a} -a \right )^{\frac {5}{6}} \left (\textit {\_a} -b \right )^{\frac {2}{3}}}d \textit {\_a} -\frac {\int _{}^{x}\left (-f \left (\textit {\_a} \right ) \left (y \left (x \right )-b \right )^{4} \left (y \left (x \right )-a \right )^{5}\right )^{\frac {1}{6}}d \textit {\_a}}{\left (y \left (x \right )-a \right )^{\frac {5}{6}} \left (y \left (x \right )-b \right )^{\frac {2}{3}}}+c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 1.947 (sec). Leaf size: 561
DSolve[(y'[x])^6 +f[x] (y[x]-a)^5 (y[x]-b)^4==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-\frac {6 \sqrt [6]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {2}{3},\frac {7}{6},\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\&\right ]\left [\int _1^x-\sqrt [6]{f(K[1])}dK[1]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [-\frac {6 \sqrt [6]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {2}{3},\frac {7}{6},\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\&\right ]\left [\int _1^x\sqrt [6]{f(K[2])}dK[2]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [-\frac {6 \sqrt [6]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {2}{3},\frac {7}{6},\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\&\right ]\left [\int _1^x-\sqrt [3]{-1} \sqrt [6]{f(K[3])}dK[3]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [-\frac {6 \sqrt [6]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {2}{3},\frac {7}{6},\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\&\right ]\left [\int _1^x\sqrt [3]{-1} \sqrt [6]{f(K[4])}dK[4]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [-\frac {6 \sqrt [6]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {2}{3},\frac {7}{6},\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\&\right ]\left [\int _1^x-(-1)^{2/3} \sqrt [6]{f(K[5])}dK[5]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [-\frac {6 \sqrt [6]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {2}{3},\frac {7}{6},\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\&\right ]\left [\int _1^x(-1)^{2/3} \sqrt [6]{f(K[6])}dK[6]+c_1\right ] \\ y(x)\to a \\ y(x)\to b \\ \end{align*}