problem 23

2.1.23 problem 23

Solved as ﬁrst order Bernoulli ode
Solved as ﬁrst order Exact ode
Maple step by step solution
Maple trace
Maple dsolve solution
Mathematica DSolve solution

Internal problem ID [8683]
Book : First order enumerated odes
Section : section 1
Problem number : 23
Date solved : Tuesday, December 17, 2024 at 12:57:44 PM
CAS classiﬁcation : [_rational, _Bernoulli]

Solve

\begin{align*} c y^{\prime }&=\frac {a x +b y^{2}}{y} \end{align*}

Solved as ﬁrst order Bernoulli ode

Time used: 0.301 (sec)

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {b \,y^{2}+a x}{y c} \end{align*}

This is a Bernoulli ODE.

\[ y' = \left (\frac {b}{c}\right ) y + \left (\frac {a x}{c}\right )\frac {1}{y} \tag {1} \]

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 &=\frac {b}{c}\\ f_1 &=\frac {a x}{c} \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)&=\frac {b}{c}\\ f_1(x)&=\frac {a x}{c}\\ n &=-1 \end{align*}

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

\begin{align*} y'y &= \frac {b \,y^{2}}{c} +\frac {a x}{c} \tag {4} \end{align*}

Let

\begin{align*} v &= y^{1-n} \\ &= y^{2} \tag {5} \end{align*}

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

\begin{align*} v' &= 2 yy' \tag {6} \end{align*}

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

\begin{align*} \frac {v^{\prime }\left (x \right )}{2}&= \frac {b v \left (x \right )}{c}+\frac {a x}{c}\\ v' &= \frac {2 b v}{c}+\frac {2 a x}{c} \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) &=-\frac {2 b}{c}\\ p(x) &=\frac {2 a x}{c} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {2 b}{c}d x}\\ &= {\mathrm e}^{-\frac {2 b x}{c}} \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 (\frac {2 a x}{c}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (v \,{\mathrm e}^{-\frac {2 b x}{c}}\right ) &= \left ({\mathrm e}^{-\frac {2 b x}{c}}\right ) \left (\frac {2 a x}{c}\right ) \\ \mathrm {d} \left (v \,{\mathrm e}^{-\frac {2 b x}{c}}\right ) &= \left (\frac {2 a x \,{\mathrm e}^{-\frac {2 b x}{c}}}{c}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} v \,{\mathrm e}^{-\frac {2 b x}{c}}&= \int {\frac {2 a x \,{\mathrm e}^{-\frac {2 b x}{c}}}{c} \,dx} \\ &=-\frac {\left (2 b x +c \right ) a \,{\mathrm e}^{-\frac {2 b x}{c}}}{2 b^{2}} + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-\frac {2 b x}{c}}\) gives the ﬁnal solution

\[ v \left (x \right ) = \frac {c_1 \,{\mathrm e}^{\frac {2 b x}{c}} b^{2}-\left (b x +\frac {c}{2}\right ) a}{b^{2}} \]

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

\[ y^{2} = \frac {c_1 \,{\mathrm e}^{\frac {2 b x}{c}} b^{2}-\left (b x +\frac {c}{2}\right ) a}{b^{2}} \]

Solving for \(y\) gives

\begin{align*} y &= -\frac {\sqrt {4 c_1 \,{\mathrm e}^{\frac {2 b x}{c}} b^{2}-4 a x b -2 a c}}{2 b} \\ y &= \frac {\sqrt {4 c_1 \,{\mathrm e}^{\frac {2 b x}{c}} b^{2}-4 a x b -2 a c}}{2 b} \\ \end{align*}

Summary of solutions found

Solved as ﬁrst order Exact ode

Time used: 0.322 (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 c\right )\mathop {\mathrm {d}y} &= \left (b \,y^{2}+a x\right )\mathop {\mathrm {d}x}\\ \left (-b \,y^{2}-a x\right )\mathop {\mathrm {d}x} + \left (y c\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

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

\begin{align*} M(x,y) &= -b \,y^{2}-a x\\ N(x,y) &= y c \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 (-b \,y^{2}-a x\right )\\ &= -2 b y \end{align*}

And

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

Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to ﬁnd an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=\frac {1}{y c}\left ( \left ( -2 b y\right ) - \left (0 \right ) \right ) \\ &=-\frac {2 b}{c} \end{align*}

Since \(A\) does not depend on \(y\), then it can be used to ﬁnd an integrating factor. The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{ \int A \mathop {\mathrm {d}x} } \\ &= e^{\int -\frac {2 b}{c}\mathop {\mathrm {d}x} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{-\frac {2 b x}{c} } \\ &= {\mathrm e}^{-\frac {2 b x}{c}} \end{align*}

\(M\) and \(N\) are multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) for now so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= {\mathrm e}^{-\frac {2 b x}{c}}\left (-b \,y^{2}-a x\right ) \\ &= -\left (b \,y^{2}+a x \right ) {\mathrm e}^{-\frac {2 b x}{c}} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= {\mathrm e}^{-\frac {2 b x}{c}}\left (y c\right ) \\ &= y c \,{\mathrm e}^{-\frac {2 b x}{c}} \end{align*}

Now a modiﬁed ODE is ontained from the original ODE, which is exact and can be solved. The modiﬁed ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (-\left (b \,y^{2}+a x \right ) {\mathrm e}^{-\frac {2 b x}{c}}\right ) + \left (y c \,{\mathrm e}^{-\frac {2 b x}{c}}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)

\begin{align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {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 \overline {M}\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -\left (b \,y^{2}+a x \right ) {\mathrm e}^{-\frac {2 b x}{c}}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \frac {c \left (2 b^{2} y^{2}+2 a x b +a c \right ) {\mathrm e}^{-\frac {2 b x}{c}}}{4 b^{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} = y c \,{\mathrm e}^{-\frac {2 b x}{c}}+f'(y) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial y} = y c \,{\mathrm e}^{-\frac {2 b x}{c}}\). Therefore equation (4) becomes

\begin{equation} \tag{5} y c \,{\mathrm e}^{-\frac {2 b x}{c}} = y c \,{\mathrm e}^{-\frac {2 b x}{c}}+f'(y) \end{equation}

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

\[ f'(y) = 0 \]

Therefore

\[ f(y) = c_1 \]

Where \(c_1\) is constant of integration. Substituting this result for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = \frac {c \left (2 b^{2} y^{2}+2 a x b +a c \right ) {\mathrm e}^{-\frac {2 b x}{c}}}{4 b^{2}}+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = \frac {c \left (2 b^{2} y^{2}+2 a x b +a c \right ) {\mathrm e}^{-\frac {2 b x}{c}}}{4 b^{2}} \]

Solving for \(y\) gives

\begin{align*} y &= -\frac {{\mathrm e}^{\frac {2 b x}{c}} \sqrt {-2 \,{\mathrm e}^{-\frac {2 b x}{c}} c \left (2 \,{\mathrm e}^{-\frac {2 b x}{c}} a b c x +{\mathrm e}^{-\frac {2 b x}{c}} a \,c^{2}-4 c_1 \,b^{2}\right )}}{2 c b} \\ y &= \frac {{\mathrm e}^{\frac {2 b x}{c}} \sqrt {-2 \,{\mathrm e}^{-\frac {2 b x}{c}} c \left (2 \,{\mathrm e}^{-\frac {2 b x}{c}} a b c x +{\mathrm e}^{-\frac {2 b x}{c}} a \,c^{2}-4 c_1 \,b^{2}\right )}}{2 c b} \\ \end{align*}

Summary of solutions found

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & c \left (\frac {d}{d x}y \left (x \right )\right )=\frac {x a +b y \left (x \right )^{2}}{y \left (x \right )} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {x a +b y \left (x \right )^{2}}{y \left (x \right ) c} \end {array} \]

Maple trace

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

Maple dsolve solution

Solving time : 0.017 (sec)
Leaf size : 69

dsolve(c*diff(y(x),x) = (a*x+b*y(x)^2)/y(x), 
       y(x),singsol=all)

\begin{align*} y &= -\frac {\sqrt {4 \,{\mathrm e}^{\frac {2 b x}{c}} c_{1} b^{2}-4 a x b -2 a c}}{2 b} \\ y &= \frac {\sqrt {4 \,{\mathrm e}^{\frac {2 b x}{c}} c_{1} b^{2}-4 a x b -2 a c}}{2 b} \\ \end{align*}

Mathematica DSolve solution

Solving time : 5.76 (sec)
Leaf size : 85

DSolve[{c*D[y[x],x]==(a*x+b*y[x]^2)/y[x],{}}, 
       y[x],x,IncludeSingularSolutions->True]

\begin{align*} y(x)\to -\frac {i \sqrt {a b x+\frac {a c}{2}+b^2 c_1 \left (-e^{\frac {2 b x}{c}}\right )}}{b} \\ y(x)\to \frac {i \sqrt {a b x+\frac {a c}{2}+b^2 c_1 \left (-e^{\frac {2 b x}{c}}\right )}}{b} \\ \end{align*}

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