Internal problem ID [2675]
Internal file name [OUTPUT/2167_Sunday_June_05_2022_02_51_25_AM_403017/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 19.
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, _Riccati]
\[ \boxed {y^{\prime } x^{2}-y^{2}-3 y x=x^{2}} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{2}+3 y x +y^{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 \(M=x^{2}+3 y x +y^{2}\) and \(N=x^{2}\) 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 &= u^{2}+3 u +1\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {u \left (x \right )^{2}+2 u \left (x \right )+1}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {u \left (x \right )^{2}+2 u \left (x \right )+1}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x -u \left (x \right )^{2}-2 u \left (x \right )-1 = 0 \] Or \[ u^{\prime }\left (x \right ) x -\left (u \left (x \right )+1\right )^{2} = 0 \] 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 {\left (u +1\right )^{2}}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=\left (u +1\right )^{2}\). Integrating both sides gives \begin{align*} \frac {1}{\left (u +1\right )^{2}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\left (u +1\right )^{2}} \,du} &= \int {\frac {1}{x} \,d x} \\ -\frac {1}{u +1}&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ -\frac {1}{u \left (x \right )+1}-\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ -\frac {1}{\frac {y}{x}+1}-\ln \left (x \right )-c_{2} = 0 \] Which simplifies to \begin {align*} -\frac {y \ln \left (x \right )+c_{2} y+\ln \left (x \right ) x +c_{2} x +x}{y+x} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {y \ln \left (x \right )+c_{2} y+\ln \left (x \right ) x +c_{2} x +x}{y+x} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {y \ln \left (x \right )+c_{2} y+\ln \left (x \right ) x +c_{2} x +x}{y+x} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x^{2}-y^{2}-3 y x =x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+3 y x +x^{2}}{x^{2}} \end {array} \]
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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 18
dsolve(x^2*diff(y(x),x)=y(x)^2+3*x*y(x)+x^2,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x \left (\ln \left (x \right )+c_{1} +1\right )}{\ln \left (x \right )+c_{1}} \]
✓ Solution by Mathematica
Time used: 0.146 (sec). Leaf size: 28
DSolve[x^2*y'[x]==y[x]^2+3*x*y[x]+x^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x (\log (x)+1+c_1)}{\log (x)+c_1} \\ y(x)\to -x \\ \end{align*}