1.59 problem 59

Internal problem ID [8023]
Book : First order enumerated odes
Section : section 1
Problem number : 59
Date solved : Monday, October 21, 2024 at 04:41:10 PM
CAS classification : [_separable]

Solve

\begin{align*} {y^{\prime }}^{2}&=\frac {1}{x^{3} y^{4}} \end{align*}

1.59.1 Solved as first order ode of type nonlinear p but separable

Time used: 0.475 (sec)

The ode has the form

\begin{align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end{align*}

Where \(n=2, m=1, f=\frac {1}{x^{3}} , g=\frac {1}{y^{4}}\). Hence the ode is

\begin{align*} (y')^{2} &= \frac {1}{x^{3} y^{4}} \end{align*}

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

\begin{align*} y^{\prime } &=\sqrt {f g}\\ y^{\prime } &=-\sqrt {f g} \end{align*}

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

\begin{align*} \frac {1}{x^{3}} &> 0\\ \frac {1}{y^{4}} &> 0 \end{align*}

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

\begin{align*} y^{\prime } &=\sqrt {f}\, \sqrt {g}\\ y^{\prime } &=-\sqrt {f}\, \sqrt {g} \end{align*}

Therefore

\begin{align*} \frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx\\ -\frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx \end{align*}

Replacing \(f(x),g(y)\) by their values gives

\begin{align*} \frac {1}{\sqrt {\frac {1}{y^{4}}}} \, dy &= \left (\sqrt {\frac {1}{x^{3}}}\right )\,dx\\ -\frac {1}{\sqrt {\frac {1}{y^{4}}}} \, dy &= \left (\sqrt {\frac {1}{x^{3}}}\right )\,dx \end{align*}

Integrating now gives the following solutions

\begin{align*} \int \frac {1}{\sqrt {\frac {1}{y^{4}}}}d y &= \int \sqrt {\frac {1}{x^{3}}}d x +c_1\\ \frac {y^{5} \sqrt {\frac {1}{y^{4}}}}{3} &= -2 x \sqrt {\frac {1}{x^{3}}}\\ \int -\frac {1}{\sqrt {\frac {1}{y^{4}}}}d y &= \int \sqrt {\frac {1}{x^{3}}}d x +c_1\\ -\frac {y^{5} \sqrt {\frac {1}{y^{4}}}}{3} &= -2 x \sqrt {\frac {1}{x^{3}}} \end{align*}

Therefore

\begin{align*} \frac {y^{5} \sqrt {\frac {1}{y^{4}}}}{3} &= -2 x \sqrt {\frac {1}{x^{3}}}+c_1 \\ -\frac {y^{5} \sqrt {\frac {1}{y^{4}}}}{3} &= -2 x \sqrt {\frac {1}{x^{3}}}+c_1 \\ \end{align*}

1.59.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}=\frac {1}{x^{3} y^{4}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {1}{x^{{3}/{2}} y^{2}}, y^{\prime }=-\frac {1}{x^{{3}/{2}} y^{2}}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {1}{x^{{3}/{2}} y^{2}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } y^{2}=\frac {1}{x^{{3}/{2}}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } y^{2}d x =\int \frac {1}{x^{{3}/{2}}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{3}}{3}=-\frac {2}{\sqrt {x}}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\left (\frac {3 \textit {\_C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {1}{x^{{3}/{2}} y^{2}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } y^{2}=-\frac {1}{x^{{3}/{2}}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } y^{2}d x =\int -\frac {1}{x^{{3}/{2}}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{3}}{3}=\frac {2}{\sqrt {x}}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\left (\frac {3 \textit {\_C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\left (\frac {3 \mathit {C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}, y=\left (\frac {3 \mathit {C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}\right \} \end {array} \]

1.59.3 Maple trace

Methods for first order ODEs:
 

1.59.4 Maple dsolve solution

Solving time : 0.030 (sec)
Leaf size : 133

dsolve(diff(y(x),x)^2 = 1/x^3/y(x)^4, 
       y(x),singsol=all)
 

\begin{align*} y &= \left (\frac {c_1 \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}} \\ y &= -\frac {\left (\frac {c_1 \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \\ y &= \frac {\left (\frac {c_1 \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{2} \\ y &= \left (\frac {c_1 \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}} \\ y &= -\frac {\left (\frac {c_1 \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \\ y &= \frac {\left (\frac {c_1 \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{2} \\ \end{align*}

1.59.5 Mathematica DSolve solution

Solving time : 3.383 (sec)
Leaf size : 157

DSolve[{(D[y[x],x])^2==1/(x^3*y[x]^4),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to -\sqrt [3]{-3} \sqrt [3]{-\frac {2}{\sqrt {x}}+c_1} \\ y(x)\to \sqrt [3]{3} \sqrt [3]{-\frac {2}{\sqrt {x}}+c_1} \\ y(x)\to (-1)^{2/3} \sqrt [3]{3} \sqrt [3]{-\frac {2}{\sqrt {x}}+c_1} \\ y(x)\to -\sqrt [3]{-3} \sqrt [3]{\frac {2}{\sqrt {x}}+c_1} \\ y(x)\to \sqrt [3]{3} \sqrt [3]{\frac {2}{\sqrt {x}}+c_1} \\ y(x)\to (-1)^{2/3} \sqrt [3]{3} \sqrt [3]{\frac {2}{\sqrt {x}}+c_1} \\ \end{align*}