Internal problem ID [5776]
Internal file name [OUTPUT/5024_Sunday_June_05_2022_03_17_55_PM_58980371/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: 28.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_linear]
\[ \boxed {y^{\prime }+\frac {x +2 y}{x}=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {x +2 y}{x}\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 \(M=-x -2 y\) and \(N=x\) are both homogeneous and of the same order \(n=1\). 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 &= -1-2 u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {-1-3 u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {-1-3 u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x +3 u \left (x \right )+1 = 0 \] 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 {-1-3 u}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=-1-3 u\). Integrating both sides gives \begin{align*} \frac {1}{-1-3 u} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{-1-3 u} \,du} &= \int {\frac {1}{x} \,d x} \\ -\frac {\ln \left (\frac {1}{3}+u \right )}{3}&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {1}{\left (\frac {1}{3}+u \right )^{{1}/{3}}} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {1}{\left (\frac {1}{3}+u \right )^{{1}/{3}}} &= c_{3} x \end {align*}
Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ y = -\frac {\left ({\mathrm e}^{3 c_{2}} c_{3}^{3} x^{3}-3\right ) {\mathrm e}^{-3 c_{2}}}{3 x^{2} c_{3}^{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left ({\mathrm e}^{3 c_{2}} c_{3}^{3} x^{3}-3\right ) {\mathrm e}^{-3 c_{2}}}{3 x^{2} c_{3}^{3}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left ({\mathrm e}^{3 c_{2}} c_{3}^{3} x^{3}-3\right ) {\mathrm e}^{-3 c_{2}}}{3 x^{2} c_{3}^{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\frac {x +2 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} \\ {} & {} & y^{\prime }=-\frac {x +2 y}{x} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} y\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & y^{\prime }=-1-\frac {2 y}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }+\frac {2 y}{x}=-1 \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+\frac {2 y}{x}\right )=-\mu \left (x \right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+\frac {2 y}{x}\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=\frac {2 \mu \left (x \right )}{x} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=x^{2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int -\mu \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int -\mu \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int -\mu \left (x \right )d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=x^{2} \\ {} & {} & y=\frac {\int -x^{2}d x +c_{1}}{x^{2}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {-\frac {x^{3}}{3}+c_{1}}{x^{2}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 13
dsolve(diff(y(x),x)+(x+2*y(x))/x=0,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x}{3}+\frac {c_{1}}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.025 (sec). Leaf size: 17
DSolve[y'[x]+(x+2*y[x])/x==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {x}{3}+\frac {c_1}{x^2} \]