1.86 problem 85

Internal problem ID [7778]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 85
Date solved : Monday, October 21, 2024 at 04:19:09 PM
CAS classification : [_Bernoulli]

Solve

\begin{align*} y^{\prime }&=2 y \left (x \sqrt {y}-1\right ) \end{align*}

With initial conditions

\begin{align*} y \left (0\right )&=1 \end{align*}

1.86.1 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)\\ &= 2 y \left (x \sqrt {y}-1\right ) \end{align*}

The \(x\) domain of \(f(x,y)\) when \(y=1\) is

\[ \{-\infty <x <\infty \} \]

And the point \(x_0 = 0\) is inside this domain. The \(y\) domain of \(f(x,y)\) when \(x=0\) is

\[ \{-\infty <y <\infty \} \]

And the point \(y_0 = 1\) is inside this domain. Now we will look at the continuity of

\begin{align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (2 y \left (x \sqrt {y}-1\right )\right ) \\ &= 3 x \sqrt {y}-2 \end{align*}

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

\[ \{-\infty <x <\infty \} \]

And the point \(x_0 = 0\) is inside this domain. The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is

\[ \{-\infty <y <\infty \} \]

And the point \(y_0 = 1\) is inside this domain. Therefore solution exists and is unique.

1.86.2 Solved as first order Bernoulli ode

Time used: 0.228 (sec)

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= 2 y \left (x \sqrt {y}-1\right ) \end{align*}

This is a Bernoulli ODE.

\[ y' = \left (-2\right ) y + \left (2 x\right )y^{{3}/{2}} \tag {1} \]

The standard Bernoulli ODE has the form

\[ y' = f_0(x)y+f_1(x)y^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=-2\\ f_1 &=2 x \end{align*}

The first step is to divide the above equation by \(y^n \) which gives

\[ \frac {y'}{y^n} = f_0(x) y^{1-n} +f_1(x) \tag {3} \]

The next step is use the substitution \(v = y^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(y(x)\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(x)&=-2\\ f_1(x)&=2 x\\ n &={\frac {3}{2}} \end{align*}

Dividing both sides of ODE (1) by \(y^n=y^{{3}/{2}}\) gives

\begin{align*} y'\frac {1}{y^{{3}/{2}}} &= -\frac {2}{\sqrt {y}} +2 x \tag {4} \end{align*}

Let

\begin{align*} v &= y^{1-n} \\ &= \frac {1}{\sqrt {y}} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(x\) gives

\begin{align*} v' &= -\frac {1}{2 y^{{3}/{2}}}y' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} -2 v^{\prime }\left (x \right )&= -2 v \left (x \right )+2 x\\ v' &= v -x \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (x \right )\) which is now solved.

In canonical form a linear first order is

\begin{align*} v^{\prime }\left (x \right ) + q(x)v \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-1\\ p(x) &=-x \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \left (-1\right )d x}\\ &= {\mathrm e}^{-x} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu v\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu v\right ) &= \left (\mu \right ) \left (-x\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (v \,{\mathrm e}^{-x}\right ) &= \left ({\mathrm e}^{-x}\right ) \left (-x\right ) \\ \mathrm {d} \left (v \,{\mathrm e}^{-x}\right ) &= \left (-x \,{\mathrm e}^{-x}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} v \,{\mathrm e}^{-x}&= \int {-x \,{\mathrm e}^{-x} \,dx} \\ &=\left (x +1\right ) {\mathrm e}^{-x} + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-x}\) gives the final solution

\[ v \left (x \right ) = c_1 \,{\mathrm e}^{x}+x +1 \]

The substitution \(v = y^{1-n}\) is now used to convert the above solution back to \(y\) which results in

\[ \frac {1}{\sqrt {y}} = c_1 \,{\mathrm e}^{x}+x +1 \]

Solving for the constant of integration from initial conditions, the solution becomes

\begin{align*} \frac {1}{\sqrt {y}} = x +1 \end{align*}

Figure 163: Slope field plot
\(y^{\prime } = 2 y \left (x \sqrt {y}-1\right )\)

1.86.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=2 y \left (x \sqrt {y}-1\right ), y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 y \left (x \sqrt {y}-1\right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]

1.86.4 Maple trace

Methods for first order ODEs:
 

1.86.5 Maple dsolve solution

Solving time : 0.020 (sec)
Leaf size : 9

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

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

1.86.6 Mathematica DSolve solution

Solving time : 0.684 (sec)
Leaf size : 20

DSolve[{D[y[x],x]==2*y[x]*(x*Sqrt[y[x]-1]),{y[0]==1}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to 1 \\ y(x)\to \sec ^2\left (\frac {x^2}{2}\right ) \\ \end{align*}