Internal problem ID [11232]
Internal file name [OUTPUT/10218_Wednesday_December_07_2022_01_21_26_PM_27725878/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter IV, differential equations of the first order and higher degree than the first.
Article 28. Summary. Page 59
Problem number: Ex 9.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "separable", "homogeneousTypeD2", "homogeneousTypeMapleC", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {{y^{\prime }}^{2} x^{2}-2 \left (x y+2 y^{\prime }\right ) y^{\prime }+y^{2}=0} \] The ode \begin {align*} {y^{\prime }}^{2} x^{2}-2 \left (x y+2 y^{\prime }\right ) y^{\prime }+y^{2} = 0 \end {align*}
is factored to \begin {align*} \left (-y^{\prime } x +y+2 y^{\prime }\right ) \left (-y^{\prime } x +y-2 y^{\prime }\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} -y^{\prime } x +y+2 y^{\prime } = 0\tag {1} \\ -y^{\prime } x +y-2 y^{\prime } = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {y}{x -2} \end {align*}
Where \(f(x)=\frac {1}{x -2}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= \frac {1}{x -2} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {\frac {1}{x -2} \,d x}\\ \ln \left (y \right )&=\ln \left (x -2\right )+c_{1}\\ y&={\mathrm e}^{\ln \left (x -2\right )+c_{1}}\\ &=c_{1} \left (x -2\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (x -2\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (x -2\right ) \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (x -2\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (x -2\right ) \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {y}{x +2} \end {align*}
Where \(f(x)=\frac {1}{x +2}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= \frac {1}{x +2} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {\frac {1}{x +2} \,d x}\\ \ln \left (y \right )&=\ln \left (x +2\right )+c_{2}\\ y&={\mathrm e}^{\ln \left (x +2\right )+c_{2}}\\ &=c_{2} \left (x +2\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} \left (x +2\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{2} \left (x +2\right ) \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} \left (x +2\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{2} \left (x +2\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2} x^{2}-2 \left (x y+2 y^{\prime }\right ) y^{\prime }+y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {y}{x -2}, y^{\prime }=\frac {y}{x +2}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y}{x -2} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{x -2} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{x -2}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\ln \left (x -2\right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1}} \left (x -2\right ) \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y}{x +2} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{x +2} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{x +2}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\ln \left (x +2\right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1}} \left (x +2\right ) \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y={\mathrm e}^{c_{1}} \left (x -2\right ), y={\mathrm e}^{c_{1}} \left (x +2\right )\right \} \end {array} \]
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: 17
dsolve(x^2*diff(y(x),x)^2-2*(x*y(x)+2*diff(y(x),x))*diff(y(x),x)+y(x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} \left (x -2\right ) \\ y \left (x \right ) &= c_{1} \left (x +2\right ) \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.074 (sec). Leaf size: 26
DSolve[x^2*(y'[x])^2-2*(x*y[x]+2*y'[x])*y'[x]+y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_1 (x-2) \\ y(x)\to c_1 (x+2) \\ y(x)\to 0 \\ \end{align*}