Internal problem ID [5773]
Internal file name [OUTPUT/5021_Sunday_June_05_2022_03_17_45_PM_40910443/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: 25.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_separable]
\[ \boxed {x^{2} {y^{\prime }}^{2}-3 x y y^{\prime }+2 y^{2}=0} \]
Solving for \(y^{\prime }\) gives \begin {align*} y^{\prime }&=\frac {y}{x}\tag {1} \\ y^{\prime }&=\frac {2 y}{x}\tag {2} \end {align*}
Now ODE (1) is solved In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {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 \(M=y\) 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 &= u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= 0 \end {align*}
Or \[ u^{\prime }\left (x \right ) = 0 \] Which is now solved as separable in \(u \left (x \right )\). Which is now solved in \(u \left (x \right )\). Integrating both sides gives \begin {align*} u \left (x \right ) &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{2} \end {align*}
Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ y = c_{2} x \] Now ODE (2) is solved In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {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 \(M=2 y\) 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 &= 2 u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x -u \left (x \right ) = 0 \] 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}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {1}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {1}{x} \,d x}\\ \ln \left (u \right )&=\ln \left (x \right )+c_{4}\\ u&={\mathrm e}^{\ln \left (x \right )+c_{4}}\\ &=x c_{4} \end {align*}
Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ y = x^{2} c_{4} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x \\ \tag{2} y &= x^{2} c_{4} \\ \end{align*}
Verification of solutions
\[ y = c_{2} x \] Verified OK.
\[ y = x^{2} c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} {y^{\prime }}^{2}-3 x y y^{\prime }+2 y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {y}{x}, y^{\prime }=\frac {2 y}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=x \,{\mathrm e}^{c_{1}} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {2 y}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {2}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {2}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=2 \ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1}} x^{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=x \,{\mathrm e}^{c_{1}}, y={\mathrm e}^{c_{1}} x^{2}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 15
dsolve(x^2*diff(y(x),x)^2-3*x*y(x)*diff(y(x),x)+2*y(x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} x^{2} \\ y \left (x \right ) &= c_{1} x \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.043 (sec). Leaf size: 24
DSolve[x^2*(y'[x])^2-3*x*y[x]*y'[x]+2*y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_1 x \\ y(x)\to c_1 x^2 \\ y(x)\to 0 \\ \end{align*}