Internal problem ID [5764]
Internal file name [OUTPUT/5012_Sunday_June_05_2022_03_17_17_PM_77468873/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: 16.
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]
\[ \boxed {x y+y^{2}-y^{\prime } x^{2}=-x^{2}} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{2}+x y +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 \(M=x^{2}+x y +y^{2}\) 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}+u +1\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {u \left (x \right )^{2}+1}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {u \left (x \right )^{2}+1}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x -u \left (x \right )^{2}-1 = 0 \] 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 {u^{2}+1}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u^{2}+1\). Integrating both sides gives \begin{align*} \frac {1}{u^{2}+1} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{u^{2}+1} \,du} &= \int {\frac {1}{x} \,d x} \\ \arctan \left (u \right )&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \arctan \left (u \left (x \right )\right )-\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \arctan \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \arctan \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \arctan \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y+y^{2}-y^{\prime } x^{2}=-x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{2}+x y+y^{2}}{x^{2}} \end {array} \]
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: 0.016 (sec). Leaf size: 11
dsolve((x^2+x*y(x)+y(x)^2)=x^2*diff(y(x),x),y(x), singsol=all)
\[ y \left (x \right ) = \tan \left (\ln \left (x \right )+c_{1} \right ) x \]
✓ Solution by Mathematica
Time used: 0.188 (sec). Leaf size: 13
DSolve[(x^2+x*y[x]+y[x]^2)==x^2*y'[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x \tan (\log (x)+c_1) \]