Internal problem ID [4065]
Internal file name [OUTPUT/3558_Sunday_June_05_2022_09_38_40_AM_14170509/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 29
Problem number: 826.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "separable", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {{y^{\prime }}^{2}-x y \left (x^{2}+y^{2}\right ) y^{\prime }+x^{4} y^{4}=0} \] The ode \begin {align*} {y^{\prime }}^{2}-x y \left (x^{2}+y^{2}\right ) y^{\prime }+x^{4} y^{4} = 0 \end {align*}
is factored to \begin {align*} \left (x^{3} y-y^{\prime }\right ) \left (x y^{3}-y^{\prime }\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} x^{3} y-y^{\prime } = 0\tag {1} \\ x y^{3}-y^{\prime } = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= x^{3} y \end {align*}
Where \(f(x)=x^{3}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= x^{3} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {x^{3} \,d x}\\ \ln \left (y \right )&=\frac {x^{4}}{4}+c_{1}\\ y&={\mathrm e}^{\frac {x^{4}}{4}+c_{1}}\\ &=c_{1} {\mathrm e}^{\frac {x^{4}}{4}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\frac {x^{4}}{4}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{\frac {x^{4}}{4}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\frac {x^{4}}{4}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{\frac {x^{4}}{4}} \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= x \,y^{3} \end {align*}
Where \(f(x)=x\) and \(g(y)=y^{3}\). Integrating both sides gives \begin{align*} \frac {1}{y^{3}} \,dy &= x \,d x \\ \int { \frac {1}{y^{3}} \,dy} &= \int {x \,d x} \\ -\frac {1}{2 y^{2}}&=\frac {x^{2}}{2}+c_{2} \\ \end{align*} Which results in \begin{align*} y &= -\frac {1}{\sqrt {-x^{2}-2 c_{2}}} \\ y &= \frac {1}{\sqrt {-x^{2}-2 c_{2}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{\sqrt {-x^{2}-2 c_{2}}} \\ \tag{2} y &= \frac {1}{\sqrt {-x^{2}-2 c_{2}}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {1}{\sqrt {-x^{2}-2 c_{2}}} \] Verified OK.
\[ y = \frac {1}{\sqrt {-x^{2}-2 c_{2}}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{\sqrt {-x^{2}-2 c_{2}}} \\ \tag{2} y &= \frac {1}{\sqrt {-x^{2}-2 c_{2}}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {1}{\sqrt {-x^{2}-2 c_{2}}} \] Verified OK.
\[ y = \frac {1}{\sqrt {-x^{2}-2 c_{2}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}-x y \left (x^{2}+y^{2}\right ) y^{\prime }+x^{4} y^{4}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=x y^{3}, y^{\prime }=x^{3} y\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=x y^{3} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{3}}=x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{3}}d x =\int x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 y^{2}}=\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {1}{\sqrt {-x^{2}-2 c_{1}}}, y=-\frac {1}{\sqrt {-x^{2}-2 c_{1}}}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=x^{3} y \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=x^{3} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int x^{3}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\frac {x^{4}}{4}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{\frac {x^{4}}{4}+c_{1}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y={\mathrm e}^{\frac {x^{4}}{4}+c_{1}}, \left \{y=\frac {1}{\sqrt {-x^{2}-2 c_{1}}}, y=-\frac {1}{\sqrt {-x^{2}-2 c_{1}}}\right \}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 38
dsolve(diff(y(x),x)^2-x*y(x)*(x^2+y(x)^2)*diff(y(x),x)+x^4*y(x)^4 = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {1}{\sqrt {-x^{2}+c_{1}}} \\ y \left (x \right ) &= -\frac {1}{\sqrt {-x^{2}+c_{1}}} \\ y \left (x \right ) &= c_{1} {\mathrm e}^{\frac {x^{4}}{4}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.188 (sec). Leaf size: 60
DSolve[(y'[x])^2-x y[x](x^2+y[x]^2)y'[x]+x^4 y[x]^4==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {1}{\sqrt {-x^2-2 c_1}} \\ y(x)\to \frac {1}{\sqrt {-x^2-2 c_1}} \\ y(x)\to c_1 e^{\frac {x^4}{4}} \\ y(x)\to 0 \\ \end{align*}