Internal problem ID [5096]
Internal file name [OUTPUT/4589_Sunday_June_05_2022_03_01_25_PM_72423458/index.tex]
Book: Engineering Mathematics. By K. A. Stroud. 5th edition. Industrial press Inc. NY.
2001
Section: Program 24. First order differential equations. Further problems 24. page
1068
Problem number: 10.
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]
\[ \boxed {\left (x^{3}+3 x y^{2}\right ) y^{\prime }-y^{3}-3 x^{2} y=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \left (3 x^{2}+y^{2}\right )}{x \left (x^{2}+3 y^{2}\right )}\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 \(M=y \left (3 x^{2}+y^{2}\right )\) and \(N=x \left (x^{2}+3 y^{2}\right )\) are both homogeneous and of the same order \(n=3\). 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 {u \left (u^{2}+3\right )}{3 u^{2}+1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {u \left (x \right ) \left (u \left (x \right )^{2}+3\right )}{3 u \left (x \right )^{2}+1}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {\frac {u \left (x \right ) \left (u \left (x \right )^{2}+3\right )}{3 u \left (x \right )^{2}+1}-u \left (x \right )}{x} = 0 \] Or \[ 3 u^{\prime }\left (x \right ) u \left (x \right )^{2} x +2 u \left (x \right )^{3}+u^{\prime }\left (x \right ) x -2 u \left (x \right ) = 0 \] Or \[ x \left (3 u \left (x \right )^{2}+1\right ) u^{\prime }\left (x \right )+2 u \left (x \right )^{3}-2 u \left (x \right ) = 0 \] 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 {2 \left (u^{3}-u \right )}{x \left (3 u^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(u)=\frac {u^{3}-u}{3 u^{2}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{3}-u}{3 u^{2}+1}} \,du &= -\frac {2}{x} \,d x \\ \int { \frac {1}{\frac {u^{3}-u}{3 u^{2}+1}} \,du} &= \int {-\frac {2}{x} \,d x} \\ -\ln \left (u \right )+2 \ln \left (u^{2}-1\right )&=-2 \ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\ln \left (u \right )+2 \ln \left (u^{2}-1\right )} &= {\mathrm e}^{-2 \ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {\left (u^{2}-1\right )^{2}}{u} &= \frac {c_{3}}{x^{2}} \end {align*}
Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ y = x \operatorname {RootOf}\left (x^{2} \textit {\_Z}^{4}-2 x^{2} \textit {\_Z}^{2}-\textit {\_Z} c_{3} +x^{2}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= x \operatorname {RootOf}\left (x^{2} \textit {\_Z}^{4}-2 x^{2} \textit {\_Z}^{2}-\textit {\_Z} c_{3} +x^{2}\right ) \\ \end{align*}
Verification of solutions
\[ y = x \operatorname {RootOf}\left (x^{2} \textit {\_Z}^{4}-2 x^{2} \textit {\_Z}^{2}-\textit {\_Z} c_{3} +x^{2}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{3}+3 x y^{2}\right ) y^{\prime }-y^{3}-3 x^{2} y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3}+3 x^{2} y}{x^{3}+3 x y^{2}} \end {array} \]
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: 0.079 (sec). Leaf size: 23
dsolve((x^3+3*x*y(x)^2)*diff(y(x),x)=y(x)^3+3*x^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {RootOf}\left (\textit {\_Z}^{4} c_{1} x -c_{1} x -\textit {\_Z} \right )^{2} x \]
✓ Solution by Mathematica
Time used: 60.142 (sec). Leaf size: 1659
DSolve[(x^3+3*x*y[x]^2)*y'[x]==y[x]^3+3*x^2*y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{6} \left (-\sqrt {3} \sqrt {4 x^2+\frac {16 \sqrt [3]{2} x^4}{\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{\sqrt [3]{2}}}-3 \sqrt {\frac {8 x^2}{3}-\frac {16 \sqrt [3]{2} x^4}{3 \sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}-\frac {2 \sqrt {3} e^{c_1} x}{\sqrt {4 x^2+\frac {16 \sqrt [3]{2} x^4}{\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{\sqrt [3]{2}}}}-\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{3 \sqrt [3]{2}}}\right ) \\ y(x)\to \frac {1}{6} \left (3 \sqrt {\frac {8 x^2}{3}-\frac {16 \sqrt [3]{2} x^4}{3 \sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}-\frac {2 \sqrt {3} e^{c_1} x}{\sqrt {4 x^2+\frac {16 \sqrt [3]{2} x^4}{\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{\sqrt [3]{2}}}}-\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{3 \sqrt [3]{2}}}-\sqrt {3} \sqrt {4 x^2+\frac {16 \sqrt [3]{2} x^4}{\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{\sqrt [3]{2}}}\right ) \\ y(x)\to \frac {1}{6} \left (\sqrt {3} \sqrt {4 x^2+\frac {16 \sqrt [3]{2} x^4}{\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{\sqrt [3]{2}}}-3 \sqrt {\frac {8 x^2}{3}-\frac {16 \sqrt [3]{2} x^4}{3 \sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {2 \sqrt {3} e^{c_1} x}{\sqrt {4 x^2+\frac {16 \sqrt [3]{2} x^4}{\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{\sqrt [3]{2}}}}-\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{3 \sqrt [3]{2}}}\right ) \\ y(x)\to \frac {1}{6} \left (\sqrt {3} \sqrt {4 x^2+\frac {16 \sqrt [3]{2} x^4}{\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{\sqrt [3]{2}}}+3 \sqrt {\frac {8 x^2}{3}-\frac {16 \sqrt [3]{2} x^4}{3 \sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {2 \sqrt {3} e^{c_1} x}{\sqrt {4 x^2+\frac {16 \sqrt [3]{2} x^4}{\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}+\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{\sqrt [3]{2}}}}-\frac {\sqrt [3]{128 x^6+27 e^{2 c_1} x^2+3 \sqrt {768 e^{2 c_1} x^8+81 e^{4 c_1} x^4}}}{3 \sqrt [3]{2}}}\right ) \\ \end{align*}