28.15 problem 813

Internal problem ID [4052]
Internal file name [OUTPUT/3545_Sunday_June_05_2022_09_37_06_AM_12870771/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 28
Problem number: 813.
ODE order: 1.
ODE degree: 2.

The type(s) of ODE detected by this program : "quadrature"

Maple gives the following as the ode type

[_quadrature]

\[ \boxed {{y^{\prime }}^{2}-2 \left (-y+x \right ) y^{\prime }-4 y x=0} \] The ode \begin {align*} {y^{\prime }}^{2}-2 \left (-y+x \right ) y^{\prime }-4 y x = 0 \end {align*}

is factored to \begin {align*} \left (y^{\prime }-2 x \right ) \left (y^{\prime }+2 y\right ) = 0 \end {align*}

Which gives the following equations \begin {align*} y^{\prime }-2 x = 0\tag {1} \\ y^{\prime }+2 y = 0\tag {2} \\ \end {align*}

Each of the above equations is now solved.

Solving ODE (1) Integrating both sides gives \begin {align*} y &= \int { 2 x\,\mathop {\mathrm {d}x}}\\ &= x^{2}+c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x^{2}+c_{1} \\ \end{align*}

Verification of solutions

\[ y = x^{2}+c_{1} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x^{2}+c_{1} \\ \end{align*}

Verification of solutions

\[ y = x^{2}+c_{1} \] Verified OK.

Solving ODE (2) Integrating both sides gives \begin {align*} \int -\frac {1}{2 y}d y &= x +c_{2}\\ -\frac {\ln \left (y \right )}{2}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&={\mathrm e}^{-2 x -2 c_{2}}\\ &=\frac {{\mathrm e}^{-2 x}}{c_{2}^{2}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-2 x}}{c_{2}^{2}} \\ \end{align*}

Verification of solutions

\[ y = \frac {{\mathrm e}^{-2 x}}{c_{2}^{2}} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-2 x}}{c_{2}^{2}} \\ \end{align*}

Verification of solutions

\[ y = \frac {{\mathrm e}^{-2 x}}{c_{2}^{2}} \] Verified OK.

28.15.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}-2 \left (-y+x \right ) y^{\prime }-4 y x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=2 x , y^{\prime }=-2 y\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=2 x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int 2 x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=x^{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=x^{2}+c_{1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-2 y \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-2 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \left (-2\right )d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-2 x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{-2 x +c_{1}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=x^{2}+c_{1} , y={\mathrm e}^{-2 x +c_{1}}\right \} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature 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: 18

dsolve(diff(y(x),x)^2-2*(x-y(x))*diff(y(x),x)-4*x*y(x) = 0,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= x^{2}+c_{1} \\ y \left (x \right ) &= {\mathrm e}^{-2 x} c_{1} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.037 (sec). Leaf size: 28

DSolve[(y'[x])^2-2(x-y[x])y'[x]-4 x y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to c_1 e^{-2 x} \\ y(x)\to x^2+c_1 \\ y(x)\to 0 \\ \end{align*}