1.23 problem 23

Internal problem ID [7987]
Book : First order enumerated odes
Section : section 1
Problem number : 23
Date solved : Monday, October 21, 2024 at 04:40:25 PM
CAS classification : [_rational, _Bernoulli]

Solve

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

1.23.1 Solved as first order Bernoulli ode

Time used: 0.353 (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 \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\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*}

1.23.2 Solved as first order Exact ode

Time used: 0.386 (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 \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\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*}

1.23.3 Maple step by step solution

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

1.23.4 Maple trace

Methods for first order ODEs:
 

1.23.5 Maple dsolve solution

Solving time : 0.006 (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 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*}

1.23.6 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*}