Internal problem ID [6212]
Internal file name [OUTPUT/5460_Sunday_June_05_2022_03_39_20_PM_57557365/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.7. Homogeneous Equations. Page
28
Problem number: 1(j).
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]
\[ \boxed {y^{3}-x y^{2} y^{\prime }=-x^{3}} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{3}+y^{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 \(N=x \,y^{2}\) 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}{u^{2}}+u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {1}{u \left (x \right )^{2} x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {1}{u \left (x \right )^{2} x} = 0 \] Or \[ u^{\prime }\left (x \right ) u \left (x \right )^{2} x -1 = 0 \] 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 {1}{u^{2} x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=\frac {1}{u^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{u^{2}}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {1}{u^{2}}} \,du} &= \int {\frac {1}{x} \,d x} \\ \frac {u^{3}}{3}&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \frac {u \left (x \right )^{3}}{3}-\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {y^{3}}{3 x^{3}}-\ln \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{3}}{3 x^{3}}-\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{3}}{3 x^{3}}-\ln \left (x \right )-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{3}-x y^{2} y^{\prime }=-x^{3} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{3}+y^{3}}{x y^{2}} \end {array} \]
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: 0.016 (sec). Leaf size: 58
dsolve((x^3+y(x)^3)-(x*y(x)^2)*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \left (c_{1} +3 \ln \left (x \right )\right )^{\frac {1}{3}} x \\ y \left (x \right ) &= -\frac {\left (c_{1} +3 \ln \left (x \right )\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right ) x}{2} \\ y \left (x \right ) &= \frac {\left (c_{1} +3 \ln \left (x \right )\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right ) x}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.197 (sec). Leaf size: 63
DSolve[(x^3+y[x]^3)-(x*y[x]^2)*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to x \sqrt [3]{3 \log (x)+c_1} \\ y(x)\to -\sqrt [3]{-1} x \sqrt [3]{3 \log (x)+c_1} \\ y(x)\to (-1)^{2/3} x \sqrt [3]{3 \log (x)+c_1} \\ \end{align*}