Internal problem ID [5766]
Internal file name [OUTPUT/5014_Sunday_June_05_2022_03_17_23_PM_2691186/index.tex]
Book: Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold
Scientific. Singapore. 1995
Section: Chapter 1. First order differential equations. Section 1.2 Homogeneous equations
problems. page 12
Problem number: 18.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _dAlembert]
\[ \boxed {y^{\prime }-\frac {2 x y}{3 x^{2}-y^{2}}=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {2 x y}{-3 x^{2}+y^{2}}\tag {1} \end {align*}
An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if \[ f(t^n x, t^n y)= t^n f(x,y) \] In this case, it can be seen that both \(M=2 x y\) and \(N=3 x^{2}-y^{2}\) are both homogeneous and of the same order \(n=2\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence \[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \] Applying the transformation \(y=ux\) to the above ODE in (1) gives \begin {align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= -\frac {2 u}{u^{2}-3}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {-\frac {2 u \left (x \right )}{u \left (x \right )^{2}-3}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {-\frac {2 u \left (x \right )}{u \left (x \right )^{2}-3}-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) u \left (x \right )^{2} x +u \left (x \right )^{3}-3 u^{\prime }\left (x \right ) x -u \left (x \right ) = 0 \] Or \[ x \left (u \left (x \right )^{2}-3\right ) u^{\prime }\left (x \right )+u \left (x \right )^{3}-u \left (x \right ) = 0 \] Which is now solved as separable in \(u \left (x \right )\). Which is now solved in \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u^{3}-u}{x \left (u^{2}-3\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{3}-u}{u^{2}-3}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{3}-u}{u^{2}-3}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{3}-u}{u^{2}-3}} \,du} &= \int {-\frac {1}{x} \,d x} \\ 3 \ln \left (u \right )-\ln \left (u^{2}-1\right )&=-\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{3 \ln \left (u \right )-\ln \left (u^{2}-1\right )} &= {\mathrm e}^{-\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {u^{3}}{u^{2}-1} &= \frac {c_{3}}{x} \end {align*}
The solution is \[ \frac {u \left (x \right )^{3}}{u \left (x \right )^{2}-1} = \frac {c_{3}}{x} \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {y^{3}}{x^{3} \left (\frac {y^{2}}{x^{2}}-1\right )} = \frac {c_{3}}{x} \] Which simplifies to \begin {align*} -\frac {y^{3}}{\left (x -y\right ) \left (y+x \right )} = c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {y^{3}}{\left (x -y\right ) \left (y+x \right )} &= c_{3} \\ \end{align*}
Verification of solutions
\[ -\frac {y^{3}}{\left (x -y\right ) \left (y+x \right )} = c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {2 x y}{3 x^{2}-y^{2}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 x y}{3 x^{2}-y^{2}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 317
dsolve(diff(y(x),x)=2*x*y(x)/(3*x^2-y(x)^2),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {1+\frac {\left (12 \sqrt {3}\, x \sqrt {27 x^{2} c_{1}^{2}-4}\, c_{1} -108 x^{2} c_{1}^{2}+8\right )^{\frac {1}{3}}}{2}+\frac {2}{\left (12 \sqrt {3}\, x \sqrt {27 x^{2} c_{1}^{2}-4}\, c_{1} -108 x^{2} c_{1}^{2}+8\right )^{\frac {1}{3}}}}{3 c_{1}} \\ y \left (x \right ) &= -\frac {\left (1+i \sqrt {3}\right ) \left (12 \sqrt {3}\, x \sqrt {27 x^{2} c_{1}^{2}-4}\, c_{1} -108 x^{2} c_{1}^{2}+8\right )^{\frac {2}{3}}-4 i \sqrt {3}-4 \left (12 \sqrt {3}\, x \sqrt {27 x^{2} c_{1}^{2}-4}\, c_{1} -108 x^{2} c_{1}^{2}+8\right )^{\frac {1}{3}}+4}{12 \left (12 \sqrt {3}\, x \sqrt {27 x^{2} c_{1}^{2}-4}\, c_{1} -108 x^{2} c_{1}^{2}+8\right )^{\frac {1}{3}} c_{1}} \\ y \left (x \right ) &= \frac {\left (i \sqrt {3}-1\right ) \left (12 \sqrt {3}\, x \sqrt {27 x^{2} c_{1}^{2}-4}\, c_{1} -108 x^{2} c_{1}^{2}+8\right )^{\frac {2}{3}}-4 i \sqrt {3}+4 \left (12 \sqrt {3}\, x \sqrt {27 x^{2} c_{1}^{2}-4}\, c_{1} -108 x^{2} c_{1}^{2}+8\right )^{\frac {1}{3}}-4}{12 \left (12 \sqrt {3}\, x \sqrt {27 x^{2} c_{1}^{2}-4}\, c_{1} -108 x^{2} c_{1}^{2}+8\right )^{\frac {1}{3}} c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.196 (sec). Leaf size: 458
DSolve[y'[x]==2*x*y[x]/(3*x^2-y[x]^2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{3} \left (\frac {\sqrt [3]{27 e^{c_1} x^2+3 \sqrt {81 e^{2 c_1} x^4-12 e^{4 c_1} x^2}-2 e^{3 c_1}}}{\sqrt [3]{2}}+\frac {\sqrt [3]{2} e^{2 c_1}}{\sqrt [3]{27 e^{c_1} x^2+3 \sqrt {81 e^{2 c_1} x^4-12 e^{4 c_1} x^2}-2 e^{3 c_1}}}-e^{c_1}\right ) \\ y(x)\to \frac {i \left (\sqrt {3}+i\right ) \sqrt [3]{27 e^{c_1} x^2+3 \sqrt {81 e^{2 c_1} x^4-12 e^{4 c_1} x^2}-2 e^{3 c_1}}}{6 \sqrt [3]{2}}-\frac {i \left (\sqrt {3}-i\right ) e^{2 c_1}}{3\ 2^{2/3} \sqrt [3]{27 e^{c_1} x^2+3 \sqrt {81 e^{2 c_1} x^4-12 e^{4 c_1} x^2}-2 e^{3 c_1}}}-\frac {e^{c_1}}{3} \\ y(x)\to -\frac {i \left (\sqrt {3}-i\right ) \sqrt [3]{27 e^{c_1} x^2+3 \sqrt {81 e^{2 c_1} x^4-12 e^{4 c_1} x^2}-2 e^{3 c_1}}}{6 \sqrt [3]{2}}+\frac {i \left (\sqrt {3}+i\right ) e^{2 c_1}}{3\ 2^{2/3} \sqrt [3]{27 e^{c_1} x^2+3 \sqrt {81 e^{2 c_1} x^4-12 e^{4 c_1} x^2}-2 e^{3 c_1}}}-\frac {e^{c_1}}{3} \\ \end{align*}