problem Problem 11

Internal problem ID [2667]
Internal ﬁle name [OUTPUT/2159_Sunday_June_05_2022_02_50_55_AM_84826546/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 1, First-Order Diﬀerential Equations. Section 1.8, Change of Variables. page 79
Problem number: Problem 11.
ODE order: 1.
ODE degree: 1.

4.3.1 Solving as homogeneous ode

In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\left (x +y \right )^{2}}{2 x^{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 \(M=\left (x +y \right )^{2}\) and \(N=2 x^{2}\) are both homogeneous and of the same order \(n=2\). 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}{2}+u +\frac {1}{2} u^{2}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {1}{2}+\frac {u \left (x \right )^{2}}{2}}{x} \end {align*}

Or \[ u^{\prime }\left (x \right )-\frac {\frac {1}{2}+\frac {u \left (x \right )^{2}}{2}}{x} = 0 \] Or \[ 2 u^{\prime }\left (x \right ) x -u \left (x \right )^{2}-1 = 0 \] 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 {\frac {u^{2}}{2}+\frac {1}{2}}{x} \end {align*}

Where \(f(x)=\frac {1}{x}\) and \(g(u)=\frac {u^{2}}{2}+\frac {1}{2}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}}{2}+\frac {1}{2}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}}{2}+\frac {1}{2}} \,du} &= \int {\frac {1}{x} \,d x} \\ 2 \arctan \left (u \right )&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ 2 \arctan \left (u \left (x \right )\right )-\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ 2 \arctan \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} = 0 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} 2 \arctan \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}

Veriﬁcation of solutions

\[ 2 \arctan \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} = 0 \] Veriﬁed OK.

4.3.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {\left (y+x \right )^{2}}{2 x^{2}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (y+x \right )^{2}}{2 x^{2}} \end {array} \]

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`

\[ y \left (x \right ) = \tan \left (\frac {\ln \left (x \right )}{2}+\frac {c_{1}}{2}\right ) x \]

4.3 problem Problem 11

4.3.1 Solving as homogeneous ode

4.3.2 Maple step by step solution