problem 1(g)

Internal problem ID [6209]
Internal ﬁle name [OUTPUT/5457_Sunday_June_05_2022_03_39_12_PM_9353929/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(g).
ODE order: 1.
ODE degree: 1.

5.7.1 Solving as homogeneous ode

In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {2 \left (-x +3 y \right )}{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 \(M=2 x -6 y\) 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 &= 2-6 u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {2-7 u \left (x \right )}{x} \end {align*}

Or \[ u^{\prime }\left (x \right )-\frac {2-7 u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x +7 u \left (x \right )-2 = 0 \] 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 {2-7 u}{x} \end {align*}

Where \(f(x)=\frac {1}{x}\) and \(g(u)=2-7 u\). Integrating both sides gives \begin{align*} \frac {1}{2-7 u} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{2-7 u} \,du} &= \int {\frac {1}{x} \,d x} \\ -\frac {\ln \left (-\frac {2}{7}+u \right )}{7}&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {1}{\left (-\frac {2}{7}+u \right )^{\frac {1}{7}}} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}

Which simpliﬁes to \begin {align*} \frac {1}{\left (-\frac {2}{7}+u \right )^{\frac {1}{7}}} &= 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 \[ y = \frac {\left (2 c_{3}^{7} {\mathrm e}^{7 c_{2}} x^{7}+7\right ) {\mathrm e}^{-7 c_{2}}}{7 x^{6} c_{3}^{7}} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (2 c_{3}^{7} {\mathrm e}^{7 c_{2}} x^{7}+7\right ) {\mathrm e}^{-7 c_{2}}}{7 x^{6} c_{3}^{7}} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {\left (2 c_{3}^{7} {\mathrm e}^{7 c_{2}} x^{7}+7\right ) {\mathrm e}^{-7 c_{2}}}{7 x^{6} c_{3}^{7}} \] Veriﬁed OK.

5.7.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x +6 y=2 x \\ \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 -6 y}{x} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} y\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & y^{\prime }=2-\frac {6 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 {6 y}{x}=2 \\ \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 {6 y}{x}\right )=2 \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 {6 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 {6 \mu \left (x \right )}{x} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=x^{6} \\ \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 2 \mu \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int 2 \mu \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int 2 \mu \left (x \right )d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=x^{6} \\ {} & {} & y=\frac {\int 2 x^{6}d x +c_{1}}{x^{6}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\frac {2 x^{7}}{7}+c_{1}}{x^{6}} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`

5.7 problem 1(g)

5.7.1 Solving as homogeneous ode

5.7.2 Maple step by step solution