2.1.86 Internal
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ID
[8224]Book
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1.0 Problem
number
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85Date
solved
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Sunday, November 10, 2024 at 03:27:38 AMCAS
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*}
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.
Time used: 0.220 (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*}
Solving for \(y\) gives
\begin{align*}
y &= \frac {1}{\left (x +1\right )^{2}} \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= \frac {1}{\left (x +1\right )^{2}} \\
\end{align*}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=2 y \left (x \right ) \left (x \sqrt {y \left (x \right )}-1\right ), y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=2 y \left (x \right ) \left (x \sqrt {y \left (x \right )}-1\right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
` Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
<- Bernoulli successful `
Solving time : 0.062
dsolve ([ diff ( y ( x ), x ) = 2*y(x)*(x*y(x)^(1/2)-1),
op ([ y (0) = 1])],y(x),singsol=all)
\[
y = \frac {1}{\left (x +1\right )^{2}}
\]
Solving time : 0.684
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*}