1.13 problem 13

Internal problem ID [5726]
Internal file name [OUTPUT/4974_Sunday_June_05_2022_03_15_39_PM_89504319/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 1.1 Separable equations problems. page 7
Problem number: 13.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "separable", "differentialType", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[_separable]

\[ \boxed {y^{\prime }-\frac {3 x^{2}+4 x +2}{2 y-2}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -1] \end {align*}

1.13.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {3 x^{2}+4 x +2}{2 y -2} \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=-1\) is \[ \{-\infty

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=-1\) is \[ \{-\infty

1.13.2 Solving as separable ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {\frac {3}{2} x^{2}+2 x +1}{y -1} \end {align*}

Where \(f(x)=\frac {3}{2} x^{2}+2 x +1\) and \(g(y)=\frac {1}{y -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y -1}} \,dy &= \frac {3}{2} x^{2}+2 x +1 \,d x \\ \int { \frac {1}{\frac {1}{y -1}} \,dy} &= \int {\frac {3}{2} x^{2}+2 x +1 \,d x} \\ \frac {y \left (y -2\right )}{2}&=\frac {1}{2} x^{3}+x^{2}+x +c_{1} \\ \end{align*} Which results in \begin{align*} y &= 1+\sqrt {x^{3}+2 x^{2}+2 c_{1} +2 x +1} \\ y &= 1-\sqrt {x^{3}+2 x^{2}+2 c_{1} +2 x +1} \\ \end{align*} Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=-1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -1 = 1-\sqrt {2 c_{1} +1} \end {align*}

The solutions are \begin {align*} c_{1} = {\frac {3}{2}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {3}{2}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=1-\sqrt {x^{3}+2 x^{2}+2 x +4} \end {align*}

The constant \(c_{1} = {\frac {3}{2}}\) gives valid solution.

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=-1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -1 = 1+\sqrt {2 c_{1} +1} \end {align*}

Warning: Unable to solve for constant of integration.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 1-\sqrt {x^{3}+2 x^{2}+2 x +4} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 1-\sqrt {x^{3}+2 x^{2}+2 x +4} \] Verified OK.

1.13.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {3 x^{2}+4 x +2}{2 y-2}=0, y \left (0\right )=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {3 x^{2}+4 x +2}{2 y-2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } \left (2 y-2\right )=3 x^{2}+4 x +2 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } \left (2 y-2\right )d x =\int \left (3 x^{2}+4 x +2\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y^{2}-2 y=x^{3}+2 x^{2}+c_{1} +2 x \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=1-\sqrt {x^{3}+2 x^{2}+c_{1} +2 x +1}, y=1+\sqrt {x^{3}+2 x^{2}+c_{1} +2 x +1}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1=1-\sqrt {1+c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =3 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =3\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\sqrt {\left (x +2\right ) \left (x^{2}+2\right )}+1 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1=1+\sqrt {1+c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\sqrt {\left (x +2\right ) \left (x^{2}+2\right )}+1 \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful`
 

✓ Solution by Maple

Time used: 0.094 (sec). Leaf size: 19

dsolve([diff(y(x),x)=(3*x^2+4*x+2)/(2*(y(x)-1)),y(0) = -1],y(x), singsol=all)
 

\[ y \left (x \right ) = 1-\sqrt {\left (x +2\right ) \left (x^{2}+2\right )} \]

✓ Solution by Mathematica

Time used: 0.132 (sec). Leaf size: 26

DSolve[{y'[x]==(3*x^2+4*x+2)/(2*(y[x]-1)),{y[0]==-1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to 1-\sqrt {x^3+2 x^2+2 x+4} \]