Internal problem ID [4200]
Internal file name [OUTPUT/3693_Sunday_June_05_2022_10_16_54_AM_1422747/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 33
Problem number: 967.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "separable", "differentialType", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {x y {y^{\prime }}^{2}+\left (3 x^{2}-2 y^{2}\right ) y^{\prime }-6 y x=0} \] The ode \begin {align*} x y {y^{\prime }}^{2}+\left (3 x^{2}-2 y^{2}\right ) y^{\prime }-6 y x = 0 \end {align*}
is factored to \begin {align*} \left (y^{\prime } y+3 x \right ) \left (-x y^{\prime }+2 y\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime } y+3 x = 0\tag {1} \\ -x y^{\prime }+2 y = 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 {3 x}{y} \end {align*}
Where \(f(x)=-3 x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -3 x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-3 x \,d x} \\ \frac {y^{2}}{2}&=-\frac {3 x^{2}}{2}+c_{1} \\ \end{align*} Which results in \begin{align*} y &= \sqrt {-3 x^{2}+2 c_{1}} \\ y &= -\sqrt {-3 x^{2}+2 c_{1}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-3 x^{2}+2 c_{1}} \\ \tag{2} y &= -\sqrt {-3 x^{2}+2 c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {-3 x^{2}+2 c_{1}} \] Verified OK.
\[ y = -\sqrt {-3 x^{2}+2 c_{1}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-3 x^{2}+2 c_{1}} \\ \tag{2} y &= -\sqrt {-3 x^{2}+2 c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {-3 x^{2}+2 c_{1}} \] Verified OK.
\[ y = -\sqrt {-3 x^{2}+2 c_{1}} \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {2 y}{x} \end {align*}
Where \(f(x)=\frac {2}{x}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= \frac {2}{x} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {\frac {2}{x} \,d x}\\ \ln \left (y \right )&=2 \ln \left (x \right )+c_{2}\\ y&={\mathrm e}^{2 \ln \left (x \right )+c_{2}}\\ &=c_{2} x^{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x^{2} \\ \end{align*}
Verification of solutions
\[ y = c_{2} x^{2} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x^{2} \\ \end{align*}
Verification of solutions
\[ y = c_{2} x^{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y {y^{\prime }}^{2}+\left (3 x^{2}-2 y^{2}\right ) y^{\prime }-6 y x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {3 x}{y}, y^{\prime }=\frac {2 y}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {3 x}{y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } y=-3 x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } yd x =\int -3 x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=-\frac {3 x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {-3 x^{2}+2 c_{1}}, y=-\sqrt {-3 x^{2}+2 c_{1}}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {2 y}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {2}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {2}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=2 \ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1}} x^{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y={\mathrm e}^{c_{1}} x^{2}, \left \{y=\sqrt {-3 x^{2}+2 c_{1}}, y=-\sqrt {-3 x^{2}+2 c_{1}}\right \}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear 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: 0.016 (sec). Leaf size: 35
dsolve(x*y(x)*diff(y(x),x)^2+(3*x^2-2*y(x)^2)*diff(y(x),x)-6*x*y(x) = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} x^{2} \\ y \left (x \right ) &= \sqrt {-3 x^{2}+c_{1}} \\ y \left (x \right ) &= -\sqrt {-3 x^{2}+c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.131 (sec). Leaf size: 54
DSolve[x y[x] (y'[x])^2+(3 x^2-2 y[x]^2)y'[x]-6 x y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_1 x^2 \\ y(x)\to -\sqrt {-3 x^2+2 c_1} \\ y(x)\to \sqrt {-3 x^2+2 c_1} \\ y(x)\to 0 \\ \end{align*}