Internal problem ID [2320]
Internal file name [OUTPUT/2320_Tuesday_February_27_2024_08_26_04_AM_83703145/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 37, page 171
Problem number: 7.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "separable", "differentialType", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {y^{2} {y^{\prime }}^{2}+y^{\prime } x y=2 x^{2}} \] The ode \begin {align*} y^{2} {y^{\prime }}^{2}+y^{\prime } x y = 2 x^{2} \end {align*}
is factored to \begin {align*} \left (y y^{\prime }+2 x \right ) \left (y y^{\prime }-x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y y^{\prime }+2 x = 0\tag {1} \\ y y^{\prime }-x = 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)\\ &= -\frac {2 x}{y} \end {align*}
Where \(f(x)=-2 x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -2 x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-2 x \,d x} \\ \frac {y^{2}}{2}&=-x^{2}+c_{1} \\ \end{align*} Which results in \begin{align*} y &= \sqrt {-2 x^{2}+2 c_{1}} \\ y &= -\sqrt {-2 x^{2}+2 c_{1}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-2 x^{2}+2 c_{1}} \\ \tag{2} y &= -\sqrt {-2 x^{2}+2 c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {-2 x^{2}+2 c_{1}} \] Verified OK.
\[ y = -\sqrt {-2 x^{2}+2 c_{1}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-2 x^{2}+2 c_{1}} \\ \tag{2} y &= -\sqrt {-2 x^{2}+2 c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {-2 x^{2}+2 c_{1}} \] Verified OK.
\[ y = -\sqrt {-2 x^{2}+2 c_{1}} \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {x}{y} \end {align*}
Where \(f(x)=x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {x \,d x} \\ \frac {y^{2}}{2}&=\frac {x^{2}}{2}+c_{2} \\ \end{align*} Which results in \begin{align*} y &= \sqrt {x^{2}+2 c_{2}} \\ y &= -\sqrt {x^{2}+2 c_{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {x^{2}+2 c_{2}} \\ \tag{2} y &= -\sqrt {x^{2}+2 c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {x^{2}+2 c_{2}} \] Verified OK.
\[ y = -\sqrt {x^{2}+2 c_{2}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {x^{2}+2 c_{2}} \\ \tag{2} y &= -\sqrt {x^{2}+2 c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {x^{2}+2 c_{2}} \] Verified OK.
\[ y = -\sqrt {x^{2}+2 c_{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2} {y^{\prime }}^{2}+y^{\prime } x y=2 x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {x}{y}, y^{\prime }=-\frac {2 x}{y}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {x}{y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y y^{\prime }=x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y y^{\prime }d x =\int x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {x^{2}+2 c_{1}}, y=-\sqrt {x^{2}+2 c_{1}}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {2 x}{y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y y^{\prime }=-2 x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y y^{\prime }d x =\int -2 x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=-x^{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {-2 x^{2}+2 c_{1}}, y=-\sqrt {-2 x^{2}+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\left \{y=\sqrt {-2 x^{2}+2 c_{1}}, y=-\sqrt {-2 x^{2}+2 c_{1}}\right \}, \left \{y=\sqrt {x^{2}+2 c_{1}}, y=-\sqrt {x^{2}+2 c_{1}}\right \}\right \} \end {array} \]
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 trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 49
dsolve(y(x)^2*diff(y(x),x)^2+x*y(x)*diff(y(x),x)-2*x^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \sqrt {x^{2}+c_{1}} \\ y \left (x \right ) &= -\sqrt {x^{2}+c_{1}} \\ y \left (x \right ) &= \sqrt {-2 x^{2}+c_{1}} \\ y \left (x \right ) &= -\sqrt {-2 x^{2}+c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.138 (sec). Leaf size: 80
DSolve[y[x]^2*y'[x]^2+x*y[x]*y'[x]-2*x^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {2} \sqrt {-x^2+c_1} \\ y(x)\to \sqrt {2} \sqrt {-x^2+c_1} \\ y(x)\to -\sqrt {x^2+2 c_1} \\ y(x)\to \sqrt {x^2+2 c_1} \\ \end{align*}