Internal problem ID [9314]
Internal file name [OUTPUT/8250_Monday_June_06_2022_02_31_20_AM_84495959/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 981.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind", "exactWithIntegrationFactor"
Maple gives the following as the ode type
[_rational, [_1st_order, `_with_symmetry_[F(x),G(x)]`], _Abel]
\[ \boxed {y^{\prime }-\frac {y^{3} a^{3} x^{3}+3 y^{2} a^{2} x^{2}+3 y a x +1+a^{2} x}{x^{3} a^{3}}=0} \]
Entering Exact first order ODE solver. (Form one type)
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 (x^{3} a^{3}\right )\mathop {\mathrm {d}y} &= \left (y^{3} a^{3} x^{3}+3 y^{2} a^{2} x^{2}+a^{2} x +3 y a x +1\right )\mathop {\mathrm {d}x}\\ \left (-y^{3} a^{3} x^{3}-3 y^{2} a^{2} x^{2}-a^{2} x -3 y a x -1\right )\mathop {\mathrm {d}x} + \left (x^{3} a^{3}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -y^{3} a^{3} x^{3}-3 y^{2} a^{2} x^{2}-a^{2} x -3 y a x -1\\ N(x,y) &= x^{3} a^{3} \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 (-y^{3} a^{3} x^{3}-3 y^{2} a^{2} x^{2}-a^{2} x -3 y a x -1\right )\\ &= -3 x a \left (y a x +1\right )^{2} \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x^{3} a^{3}\right )\\ &= 3 x^{2} a^{3} \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}{x^{3} a^{3}}\left ( \left ( -3 a^{3} x^{3} y^{2}-6 a^{2} x^{2} y -3 x a\right ) - \left (3 x^{2} a^{3} \right ) \right ) \\ &=\frac {-3+\left (-3 x^{2} y^{2}-3 x \right ) a^{2}-6 y a x}{a^{2} x^{2}} \end {align*}
Since \(A\) depends on \(y\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let \begin {align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} \right ) \\ &=-\frac {1}{1+y^{3} a^{3} x^{3}+\left (3 x^{2} y^{2}+x \right ) a^{2}+3 y a x}\left ( \left ( 3 x^{2} a^{3}\right ) - \left (-3 a^{3} x^{3} y^{2}-6 a^{2} x^{2} y -3 x a \right ) \right ) \\ &=-\frac {3 a x \left (1+\left (x^{2} y^{2}+x \right ) a^{2}+2 y a x \right )}{1+y^{3} a^{3} x^{3}+\left (3 x^{2} y^{2}+x \right ) a^{2}+3 y a x} \end {align*}
Since \(B\) depends on \(x\), it can not be used to obtain an integrating factor.We will now try a third method to find an integrating factor. Let \[ R = \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \] \(R\) is now checked to see if it is a function of only \(t=xy\). Therefore \begin {align*} R &= \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \\ &= \frac {\left (3 x^{2} a^{3}\right )-\left (-3 a^{3} x^{3} y^{2}-6 a^{2} x^{2} y -3 x a\right )} {x\left (-y^{3} a^{3} x^{3}-3 y^{2} a^{2} x^{2}-a^{2} x -3 y a x -1\right ) - y\left (x^{3} a^{3}\right )} \\ &= -\frac {3 a}{y a x +1} \end {align*}
Replacing all powers of terms \(xy\) by \(t\) gives \[ R = -\frac {3 a}{a t +1} \] Since \(R\) depends on \(t\) only, then it can be used to find an integrating factor. Let the integrating factor be \(\mu \) then \begin {align*} \mu &= e^{\int R \mathop {\mathrm {d}t}} \\ &= e^{\int \left (-\frac {3 a}{a t +1}\right )\mathop {\mathrm {d}t} } \end {align*}
The result of integrating gives \begin {align*} \mu &= e^{-3 \ln \left (a t +1\right ) } \\ &= \frac {1}{\left (a t +1\right )^{3}} \end {align*}
Now \(t\) is replaced back with \(xy\) giving \[ \mu =\frac {1}{\left (y a x +1\right )^{3}} \] Multiplying \(M\) and \(N\) by this integrating factor gives new \(M\) and new \(N\) which are called \( \overline {M}\) and \( \overline {N}\) so not to confuse them with the original \(M\) and \(N\) \begin {align*} \overline {M} &=\mu M \\ &= \frac {1}{\left (y a x +1\right )^{3}}\left (-y^{3} a^{3} x^{3}-3 y^{2} a^{2} x^{2}-a^{2} x -3 y a x -1\right ) \\ &= \frac {-1-y^{3} a^{3} x^{3}+\left (-3 x^{2} y^{2}-x \right ) a^{2}-3 y a x}{\left (y a x +1\right )^{3}} \end {align*}
And \begin {align*} \overline {N} &=\mu N \\ &= \frac {1}{\left (y a x +1\right )^{3}}\left (x^{3} a^{3}\right ) \\ &= \frac {x^{3} a^{3}}{\left (y a x +1\right )^{3}} \end {align*}
A modified ODE is now obtained from the original ODE, which is exact and can solved. The modified ODE is \begin {align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (\frac {-1-y^{3} a^{3} x^{3}+\left (-3 x^{2} y^{2}-x \right ) a^{2}-3 y a x}{\left (y a x +1\right )^{3}}\right ) + \left (\frac {x^{3} a^{3}}{\left (y a x +1\right )^{3}}\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 \frac {-1-y^{3} a^{3} x^{3}+\left (-3 x^{2} y^{2}-x \right ) a^{2}-3 y a x}{\left (y a x +1\right )^{3}}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= -x -\frac {1}{2 y^{2} \left (y a x +1\right )^{2}}+\frac {1}{y^{2} \left (y a x +1\right )}+ 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{align*} \tag{4} \frac {\partial \phi }{\partial y} &= \frac {1}{y^{3} \left (y a x +1\right )^{2}}+\frac {x a}{y^{2} \left (y a x +1\right )^{3}}-\frac {2}{y^{3} \left (y a x +1\right )}-\frac {x a}{y^{2} \left (y a x +1\right )^{2}}+f'(y) \\ &=\frac {-3 y^{2} a^{2} x^{2}-3 y a x -1}{y^{3} \left (y a x +1\right )^{3}}+f'(y) \\ \end{align*} But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {x^{3} a^{3}}{\left (y a x +1\right )^{3}}\). Therefore equation (4) becomes \begin{equation} \tag{5} \frac {x^{3} a^{3}}{\left (y a x +1\right )^{3}} = \frac {-3 y^{2} a^{2} x^{2}-3 y a x -1}{y^{3} \left (y a x +1\right )^{3}}+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = \frac {1}{y^{3}} \] Integrating the above w.r.t \(y\) gives \begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {1}{y^{3}}\right ) \mathop {\mathrm {d}y} \\ f(y) &= -\frac {1}{2 y^{2}}+ c_{1} \\ \end{align*} Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = -x -\frac {1}{2 y^{2} \left (y a x +1\right )^{2}}+\frac {1}{y^{2} \left (y a x +1\right )}-\frac {1}{2 y^{2}}+ c_{1} \] 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 new constant \(c_{1}\) gives the solution as \[ c_{1} = -x -\frac {1}{2 y^{2} \left (y a x +1\right )^{2}}+\frac {1}{y^{2} \left (y a x +1\right )}-\frac {1}{2 y^{2}} \]
The solution(s) found are the following \begin{align*} \tag{1} -x -\frac {1}{2 y^{2} \left (y a x +1\right )^{2}}+\frac {1}{y^{2} \left (y a x +1\right )}-\frac {1}{2 y^{2}} &= c_{1} \\ \end{align*}
Verification of solutions
\[ -x -\frac {1}{2 y^{2} \left (y a x +1\right )^{2}}+\frac {1}{y^{2} \left (y a x +1\right )}-\frac {1}{2 y^{2}} = c_{1} \] Verified OK.
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=y^{3}+\frac {3 y^{2}}{x a}+\frac {3 y}{a^{2} x^{2}}+\frac {a^{2} x +1}{x^{3} a^{3}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {1}{x^{2} a}+\frac {1}{x^{3} a^{3}}\\ f_1(x) &= \frac {3}{a^{2} x^{2}}\\ f_2(x) &= \frac {3}{x a}\\ f_3(x) &= 1 \end {align*}
Since \(f_2(x)=\frac {3}{x a}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {\frac {3}{x a}}{3} \right ) \\ &= u \left (x \right )-\frac {1}{x a} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = u \left (x \right )^{3}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
Integrating both sides gives \begin {align*} \int \frac {1}{u^{3}}d u &= x +c_{2}\\ -\frac {1}{2 u^{2}}&=x +c_{2} \end {align*}
Solving for \(u\) gives these solutions \begin {align*} u_1&=\frac {1}{\sqrt {-2 c_{2} -2 x}}\\ u_2&=-\frac {1}{\sqrt {-2 c_{2} -2 x}} \end {align*}
Now we transform the solution \(u \left (x \right ) = \frac {1}{\sqrt {-2 c_{2} -2 x}}\) to \(y\) using \(y = u(x) - \frac {f_2}{3 f_3}\), which gives \begin {align*} y &= \frac {1}{\sqrt {-2 c_{2} -2 x}} - \left (\frac {1}{x a}\right ) \\ &= \frac {1}{\sqrt {-2 c_{2} -2 x}}-\frac {3}{x a}\\ &= \frac {1}{\sqrt {-2 c_{2} -2 x}}-\frac {3}{x a} \end {align*}
Now we transform the solution \(u \left (x \right ) = -\frac {1}{\sqrt {-2 c_{2} -2 x}}\) to \(y\) using \(y = u(x) - \frac {f_2}{3 f_3}\), which gives \begin {align*} y &= -\frac {1}{\sqrt {-2 c_{2} -2 x}} - \left (\frac {1}{x a}\right ) \\ &= -\frac {1}{\sqrt {-2 c_{2} -2 x}}-\frac {3}{x a}\\ &= -\frac {1}{\sqrt {-2 c_{2} -2 x}}-\frac {3}{x a} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {1}{\sqrt {-2 c_{2} -2 x}}-\frac {3}{x a} \\ \tag{2} y &= -\frac {1}{\sqrt {-2 c_{2} -2 x}}-\frac {3}{x a} \\ \end{align*}
Verification of solutions
\[ y = \frac {1}{\sqrt {-2 c_{2} -2 x}}-\frac {3}{x a} \] Verified OK.
\[ y = -\frac {1}{\sqrt {-2 c_{2} -2 x}}-\frac {3}{x a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {y^{3} a^{3} x^{3}+3 y^{2} a^{2} x^{2}+3 y a x +1+a^{2} x}{x^{3} a^{3}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3} a^{3} x^{3}+3 y^{2} a^{2} x^{2}+3 y a x +1+a^{2} x}{x^{3} a^{3}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 41
dsolve(diff(y(x),x) = (y(x)^3*a^3*x^3+3*y(x)^2*a^2*x^2+3*y(x)*a*x+1+a^2*x)/x^3/a^3,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\frac {1}{\sqrt {-2 x +c_{1}}}-\frac {1}{a x} \\ y \left (x \right ) &= \frac {1}{\sqrt {-2 x +c_{1}}}-\frac {1}{a x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.29 (sec). Leaf size: 61
DSolve[y'[x] == (1 + a^2*x + 3*a*x*y[x] + 3*a^2*x^2*y[x]^2 + a^3*x^3*y[x]^3)/(a^3*x^3),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {1}{a x}-\frac {1}{\sqrt {-2 x+c_1}} \\ y(x)\to -\frac {1}{a x}+\frac {1}{\sqrt {-2 x+c_1}} \\ y(x)\to -\frac {1}{a x} \\ \end{align*}