1.2 problem 2

Internal problem ID [3797]
Book : Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section : Chapter 2. First-Order and Simple Higher-Order Differential Equations. Page 78
Problem number : 2
Date solved : Thursday, October 17, 2024 at 04:39:58 AM
CAS classification : [_separable]

Solve

\begin{align*} y^{\prime }&=\frac {x^{3} {\mathrm e}^{x^{2}}}{y \ln \left (y\right )} \end{align*}

1.2.1 Solved as first order Exact ode

Time used: 0.259 (sec)

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]

Therefore

\begin{align*} \left (y \ln \left (y \right )\right )\mathop {\mathrm {d}y} &= \left (x^{3} {\mathrm e}^{x^{2}}\right )\mathop {\mathrm {d}x}\\ \left (-x^{3} {\mathrm e}^{x^{2}}\right )\mathop {\mathrm {d}x} + \left (y \ln \left (y \right )\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(x,y) &= -x^{3} {\mathrm e}^{x^{2}}\\ N(x,y) &= y \ln \left (y \right ) \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-x^{3} {\mathrm e}^{x^{2}}\right )\\ &= 0 \end{align*}

And

\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (y \ln \left (y \right )\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)

\begin{align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end{align*}

Integrating (1) w.r.t. \(x\) gives

\begin{align*} \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int M\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -x^{3} {\mathrm e}^{x^{2}}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= -\frac {\left (x^{2}-1\right ) {\mathrm e}^{x^{2}}}{2}+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial y} = y \ln \left (y \right )\). Therefore equation (4) becomes

\begin{equation} \tag{5} y \ln \left (y \right ) = 0+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = y \ln \left (y \right ) \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( y \ln \left (y \right )\right ) \mathop {\mathrm {d}y} \\ f(y) &= \frac {y^{2} \ln \left (y \right )}{2}-\frac {y^{2}}{4}+ c_2 \\ \end{align*}

Where \(c_2\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = -\frac {\left (x^{2}-1\right ) {\mathrm e}^{x^{2}}}{2}+\frac {y^{2} \ln \left (y \right )}{2}-\frac {y^{2}}{4}+ c_2 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_2\) and \(c_2\) constants into the constant \(c_2\) gives the solution as

\[ c_2 = -\frac {\left (x^{2}-1\right ) {\mathrm e}^{x^{2}}}{2}+\frac {y^{2} \ln \left (y \right )}{2}-\frac {y^{2}}{4} \]

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y = {\mathrm e}^{\frac {\operatorname {LambertW}\left (2 \left (x^{2} {\mathrm e}^{x^{2}}-{\mathrm e}^{x^{2}}+2 c_2 \right ) {\mathrm e}^{-1}\right )}{2}+\frac {1}{2}} \end{align*}

Figure 2: Slope field plot
\(y^{\prime } = \frac {x^{3} {\mathrm e}^{x^{2}}}{y \ln \left (y\right )}\)

1.2.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {x^{3} {\mathrm e}^{x^{2}}}{y \left (x \right ) \ln \left (y \left (x \right )\right )} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {x^{3} {\mathrm e}^{x^{2}}}{y \left (x \right ) \ln \left (y \left (x \right )\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) y \left (x \right ) \ln \left (y \left (x \right )\right )=x^{3} {\mathrm e}^{x^{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right ) y \left (x \right ) \ln \left (y \left (x \right )\right )d x =\int x^{3} {\mathrm e}^{x^{2}}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y \left (x \right )^{2} \ln \left (y \left (x \right )\right )}{2}-\frac {y \left (x \right )^{2}}{4}=\frac {\left (x^{2}-1\right ) {\mathrm e}^{x^{2}}}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )={\mathrm e}^{\frac {\mathit {LambertW}\left (\frac {2 \left (x^{2} {\mathrm e}^{x^{2}}-{\mathrm e}^{x^{2}}+2 \mathit {C1} \right )}{{\mathrm e}}\right )}{2}+\frac {1}{2}} \end {array} \]

1.2.3 Maple trace

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

1.2.4 Maple dsolve solution

Solving time : 0.035 (sec)
Leaf size : 54

dsolve(diff(y(x),x) = x^3*exp(x^2)/y(x)/ln(y(x)), 
       y(x),singsol=all)
 

\[ y \left (x \right ) = \sqrt {2}\, \sqrt {\frac {x^{2} {\mathrm e}^{x^{2}}-{\mathrm e}^{x^{2}}+2 c_{1}}{\operatorname {LambertW}\left (2 \left (x^{2} {\mathrm e}^{x^{2}}-{\mathrm e}^{x^{2}}+2 c_{1} \right ) {\mathrm e}^{-1}\right )}} \]

1.2.5 Mathematica DSolve solution

Solving time : 60.187 (sec)
Leaf size : 106

DSolve[{D[y[x],x]==(x^3*Exp[x^2])/(y[x]*Log[y[x]]),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to -\frac {\sqrt {2 e^{x^2} \left (x^2-1\right )+4 c_1}}{\sqrt {W\left (\frac {2 e^{x^2} \left (x^2-1\right )+4 c_1}{e}\right )}} \\ y(x)\to \frac {\sqrt {2 e^{x^2} \left (x^2-1\right )+4 c_1}}{\sqrt {W\left (\frac {2 e^{x^2} \left (x^2-1\right )+4 c_1}{e}\right )}} \\ \end{align*}