Internal problem ID [5094]
Internal file name [OUTPUT/4587_Sunday_June_05_2022_03_01_23_PM_6639194/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: 8.
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, _Bernoulli]
\[ \boxed {y^{3}-3 x y^{2} y^{\prime }=-x^{3}} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{3}+y^{3}}{3 x \,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 \(M=x^{3}+y^{3}\) and \(N=3 y^{2} x\) 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 {1}{3 u^{2}}+\frac {u}{3}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {1}{3 u \left (x \right )^{2}}-\frac {2 u \left (x \right )}{3}}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {\frac {1}{3 u \left (x \right )^{2}}-\frac {2 u \left (x \right )}{3}}{x} = 0 \] Or \[ 3 u^{\prime }\left (x \right ) u \left (x \right )^{2} x +2 u \left (x \right )^{3}-1 = 0 \] 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 u^{3}-1}{3 u^{2} x} \end {align*}
Where \(f(x)=-\frac {1}{3 x}\) and \(g(u)=\frac {2 u^{3}-1}{u^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {2 u^{3}-1}{u^{2}}} \,du &= -\frac {1}{3 x} \,d x \\ \int { \frac {1}{\frac {2 u^{3}-1}{u^{2}}} \,du} &= \int {-\frac {1}{3 x} \,d x} \\ \frac {\ln \left (u^{3}-\frac {1}{2}\right )}{6}&=-\frac {\ln \left (x \right )}{3}+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {2^{\frac {2}{3}} \left (4 u^{3}-2\right )^{\frac {1}{6}}}{2} &= {\mathrm e}^{-\frac {\ln \left (x \right )}{3}+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {2^{\frac {2}{3}} \left (4 u^{3}-2\right )^{\frac {1}{6}}}{2} &= \frac {c_{3}}{x^{\frac {1}{3}}} \end {align*}
Which simplifies to \[ \frac {2^{\frac {2}{3}} \left (4 u \left (x \right )^{3}-2\right )^{\frac {1}{6}}}{2} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x^{\frac {1}{3}}} \] The solution is \[ \frac {2^{\frac {2}{3}} \left (4 u \left (x \right )^{3}-2\right )^{\frac {1}{6}}}{2} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x^{\frac {1}{3}}} \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {2^{\frac {2}{3}} \left (\frac {4 y^{3}}{x^{3}}-2\right )^{\frac {1}{6}}}{2} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x^{\frac {1}{3}}} \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {2^{\frac {5}{6}} \left (\frac {2 y^{3}-x^{3}}{x^{3}}\right )^{\frac {1}{6}}}{2} &= \frac {c_{3} {\mathrm e}^{c_{2}}}{x^{\frac {1}{3}}} \\ \end{align*}
Verification of solutions
\[ \frac {2^{\frac {5}{6}} \left (\frac {2 y^{3}-x^{3}}{x^{3}}\right )^{\frac {1}{6}}}{2} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x^{\frac {1}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{3}-3 x y^{2} y^{\prime }=-x^{3} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{3}+y^{3}}{3 x y^{2}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 74
dsolve((x^3+y(x)^3)=3*x*y(x)^2*diff(y(x),x),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {2^{\frac {2}{3}} {\left (x \left (x^{2}+2 c_{1} \right )\right )}^{\frac {1}{3}}}{2} \\ y \left (x \right ) &= -\frac {2^{\frac {2}{3}} {\left (x \left (x^{2}+2 c_{1} \right )\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}{4} \\ y \left (x \right ) &= \frac {2^{\frac {2}{3}} {\left (x \left (x^{2}+2 c_{1} \right )\right )}^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}{4} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.21 (sec). Leaf size: 90
DSolve[(x^3+y[x]^3)==3*x*y[x]^2*y'[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt [3]{-\frac {1}{2}} \sqrt [3]{x} \sqrt [3]{x^2+2 c_1} \\ y(x)\to \frac {\sqrt [3]{x} \sqrt [3]{x^2+2 c_1}}{\sqrt [3]{2}} \\ y(x)\to \frac {(-1)^{2/3} \sqrt [3]{x} \sqrt [3]{x^2+2 c_1}}{\sqrt [3]{2}} \\ \end{align*}