problem 79

Dividing throughout by the integrating factor \({\mathrm e}^{-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+a^{2}}\right )}{\sqrt {a^{2}}}-\frac {\ln \left (x^{2}+1\right )}{2}}\) gives the ﬁnal solution

1.80.2 Solved as ﬁrst order separable ode

The ode \(y^{\prime } = \frac {y \left (a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}\right )}{a^{2} \left (x^{2}+1\right )}\) is separable as it can be written as

\begin{align*} y^{\prime }&= \frac {y \left (a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}\right )}{a^{2} \left (x^{2}+1\right )}\\ &= f(x) g(y) \end{align*}

\begin{align*} f(x) &= \frac {a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}}{a^{2} \left (x^{2}+1\right )}\\ g(y) &= y \end{align*}

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { \frac {1}{y}\,dy} &= \int { \frac {a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}}{a^{2} \left (x^{2}+1\right )} \,dx}\\ \ln \left (y\right )&=\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+a^{2}}\right )}{\sqrt {a^{2}}}+\ln \left (\sqrt {x^{2}+1}\right )+c_1 \end{align*}

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(y)\) is zero, since we had to divide by this above. Solving \(g(y)=0\) or \(y=0\) for \(y\) gives

\begin{align*} y&=0 \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

\begin{align*} \ln \left (y\right ) = \frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+a^{2}}\right )}{\sqrt {a^{2}}}+\ln \left (\sqrt {x^{2}+1}\right )+c_1\\ y = 0 \end{align*}

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

\begin{align*} y&=0\\ y&={\mathrm e}^{\frac {2 \ln \left (\sqrt {x^{2}+1}\right ) \sqrt {a^{2}}+2 c_1 \sqrt {a^{2}}+2 \ln \left (a^{2} x +\sqrt {a^{2} x^{2}+a^{2}}\, \sqrt {a^{2}}\right )-\ln \left (a^{2}\right )}{2 \sqrt {a^{2}}}} \end{align*}

1.80.3 Solved as ﬁrst order homogeneous class D2 ode

Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes

\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = \frac {u \left (x \right ) x \left (1+\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )}{\sqrt {a^{2} \left (x^{2}+1\right )}} \end{align*}

Which is now solved The ode \(u^{\prime }\left (x \right ) = \frac {u \left (x \right ) \left (x \sqrt {a^{2} \left (x^{2}+1\right )}-a^{2}\right )}{a^{2} \left (x^{2}+1\right ) x}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right ) \left (x \sqrt {a^{2} \left (x^{2}+1\right )}-a^{2}\right )}{a^{2} \left (x^{2}+1\right ) x}\\ &= f(x) g(u) \end{align*}

\begin{align*} f(x) &= \frac {x \sqrt {a^{2} \left (x^{2}+1\right )}-a^{2}}{\left (x^{2}+1\right ) x \,a^{2}}\\ g(u) &= u \end{align*}

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{u}\,du} &= \int { \frac {x \sqrt {a^{2} \left (x^{2}+1\right )}-a^{2}}{\left (x^{2}+1\right ) x \,a^{2}} \,dx}\\ \ln \left (u \left (x \right )\right )&=\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+a^{2}}\right )}{\sqrt {a^{2}}}+\ln \left (\frac {\sqrt {x^{2}+1}}{x}\right )+c_1 \end{align*}

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(u=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0 \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

\begin{align*} \ln \left (u \left (x \right )\right ) = \frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+a^{2}}\right )}{\sqrt {a^{2}}}+\ln \left (\frac {\sqrt {x^{2}+1}}{x}\right )+c_1\\ u \left (x \right ) = 0 \end{align*}

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

\begin{align*} u \left (x \right )&=0\\ u \left (x \right )&={\mathrm e}^{\frac {2 \sqrt {a^{2}}\, \ln \left (\frac {1}{x}\right )+\ln \left (x^{2}+1\right ) \sqrt {a^{2}}+2 c_1 \sqrt {a^{2}}+2 \ln \left (a^{2} x +\sqrt {a^{2} x^{2}+a^{2}}\, \sqrt {a^{2}}\right )-\ln \left (a^{2}\right )}{2 \sqrt {a^{2}}}} \end{align*}

\begin{align*} y = 0 \end{align*}

Converting \(u \left (x \right ) = {\mathrm e}^{\frac {2 \sqrt {a^{2}}\, \ln \left (\frac {1}{x}\right )+\ln \left (x^{2}+1\right ) \sqrt {a^{2}}+2 c_1 \sqrt {a^{2}}+2 \ln \left (a^{2} x +\sqrt {a^{2} x^{2}+a^{2}}\, \sqrt {a^{2}}\right )-\ln \left (a^{2}\right )}{2 \sqrt {a^{2}}}}\) back to \(y\) gives

\begin{align*} y = x \,{\mathrm e}^{\frac {2 \sqrt {a^{2}}\, \ln \left (\frac {1}{x}\right )+\ln \left (x^{2}+1\right ) \sqrt {a^{2}}+2 c_1 \sqrt {a^{2}}+2 \ln \left (a^{2} x +\sqrt {a^{2} x^{2}+a^{2}}\, \sqrt {a^{2}}\right )-\ln \left (a^{2}\right )}{2 \sqrt {a^{2}}}} \end{align*}

1.80.4 Solved as ﬁrst order Exact ode

\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

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

\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

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

\begin{align*} \mathop {\mathrm {d}y} &= \left (\frac {y \left (1+\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )\mathop {\mathrm {d}x}\\ \left (-\frac {y \left (1+\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right ) \mathop {\mathrm {d}x} + \mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

\begin{align*} M(x,y) &= -\frac {y \left (1+\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )}{\sqrt {a^{2} \left (x^{2}+1\right )}}\\ N(x,y) &= 1 \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

\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-\frac {y \left (1+\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )\\ &= \frac {-a^{2} x -\sqrt {a^{2} \left (x^{2}+1\right )}}{a^{2} \left (x^{2}+1\right )} \end{align*}

\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (1\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 ) \\ &=1\left ( \left ( -\frac {1+\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right ) - \left (0 \right ) \right ) \\ &=\frac {-a^{2} x -\sqrt {a^{2} \left (x^{2}+1\right )}}{a^{2} \left (x^{2}+1\right )} \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 {-a^{2} x -\sqrt {a^{2} \left (x^{2}+1\right )}}{a^{2} \left (x^{2}+1\right )}\mathop {\mathrm {d}x} } \end{align*}

\begin{align*} \mu &= e^{-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+a^{2}}\right )}{\sqrt {a^{2}}}-\frac {\ln \left (x^{2}+1\right )}{2} } \\ &= \frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{\sqrt {x^{2}+1}} \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 \\ &= \frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{\sqrt {x^{2}+1}}\left (-\frac {y \left (1+\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right ) \\ &= \frac {\left (-a^{2} x -\sqrt {a^{2} \left (x^{2}+1\right )}\right ) y \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{a^{2} \left (x^{2}+1\right )^{{3}/{2}}} \end{align*}

\begin{align*} \overline {N} &=\mu N \\ &= \frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{\sqrt {x^{2}+1}}\left (1\right ) \\ &= \frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{\sqrt {x^{2}+1}} \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 (\frac {\left (-a^{2} x -\sqrt {a^{2} \left (x^{2}+1\right )}\right ) y \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{a^{2} \left (x^{2}+1\right )^{{3}/{2}}}\right ) + \left (\frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{\sqrt {x^{2}+1}}\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*}

\begin{align*} \int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int \overline {N}\mathop {\mathrm {d}y} \\ \int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int \frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{\sqrt {x^{2}+1}}\mathop {\mathrm {d}y} \\ \tag{3} \phi &= \frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} y}{\sqrt {x^{2}+1}}+ f(x) \\ \end{align*}

Where \(f(x)\) 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 \(x\) gives

\begin{align*} \tag{4} \frac {\partial \phi }{\partial x} &= -\frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} \operatorname {csgn}\left (a \right ) \left (\operatorname {csgn}\left (a \right ) a +\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right ) y}{a \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right ) \sqrt {x^{2}+1}}-\frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} y x}{\left (x^{2}+1\right )^{{3}/{2}}}+f'(x) \\ &=-\frac {y \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} \left (\left (\operatorname {csgn}\left (a \right ) a \,x^{2}+x^{2}+1\right ) \sqrt {a^{2} \left (x^{2}+1\right )}+a x \left (x^{2}+1\right ) \left (a +\operatorname {csgn}\left (a \right )\right )\right )}{\left (x^{2}+1\right )^{{3}/{2}} \sqrt {a^{2} \left (x^{2}+1\right )}}+f'(x) \\ \end{align*}

But equation (1) says that \(\frac {\partial \phi }{\partial x} = \frac {\left (-a^{2} x -\sqrt {a^{2} \left (x^{2}+1\right )}\right ) y \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{a^{2} \left (x^{2}+1\right )^{{3}/{2}}}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {\left (-a^{2} x -\sqrt {a^{2} \left (x^{2}+1\right )}\right ) y \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}}{a^{2} \left (x^{2}+1\right )^{{3}/{2}}} = -\frac {y \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} \left (\left (\operatorname {csgn}\left (a \right ) a \,x^{2}+x^{2}+1\right ) \sqrt {a^{2} \left (x^{2}+1\right )}+a x \left (x^{2}+1\right ) \left (a +\operatorname {csgn}\left (a \right )\right )\right )}{\left (x^{2}+1\right )^{{3}/{2}} \sqrt {a^{2} \left (x^{2}+1\right )}}+f'(x) \end{equation}

\begin{align*} f'(x) &= \frac {y \left (\operatorname {csgn}\left (a \right ) \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} \sqrt {a^{2} \left (x^{2}+1\right )}\, a \,x^{2}+\operatorname {csgn}\left (a \right ) \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} a \,x^{3}+\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} a^{2} x^{3}+\operatorname {csgn}\left (a \right ) \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} a x +\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} \sqrt {a^{2} \left (x^{2}+1\right )}\, x^{2}+\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} a^{2} x -\sqrt {a^{2} \left (x^{2}+1\right )}\, \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} x -\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} x^{2}+\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}} \sqrt {a^{2} \left (x^{2}+1\right )}-\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}}\right )}{\left (x^{2}+1\right )^{{3}/{2}} \sqrt {a^{2} \left (x^{2}+1\right )}} \\ &= \frac {\left (\left (\left (\operatorname {csgn}\left (a \right ) a \,x^{2}+x^{2}+1\right ) \sqrt {a^{2} \left (x^{2}+1\right )}+a x \left (x^{2}+1\right ) \left (a +\operatorname {csgn}\left (a \right )\right )\right ) \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}}-\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} \left (x \sqrt {a^{2} \left (x^{2}+1\right )}+x^{2}+1\right )\right ) y}{\left (x^{2}+1\right )^{{3}/{2}} \sqrt {a^{2} \left (x^{2}+1\right )}}\\ \end{align*}

\begin{align*} \int f'(x) \mathop {\mathrm {d}x} &= \int \left ( \frac {\left (\left (\left (\operatorname {csgn}\left (a \right ) a \,x^{2}+x^{2}+1\right ) \sqrt {a^{2} \left (x^{2}+1\right )}+a x \left (x^{2}+1\right ) \left (a +\operatorname {csgn}\left (a \right )\right )\right ) \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}}-\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} \left (x \sqrt {a^{2} \left (x^{2}+1\right )}+x^{2}+1\right )\right ) y}{\left (x^{2}+1\right )^{{3}/{2}} \sqrt {a^{2} \left (x^{2}+1\right )}}\right ) \mathop {\mathrm {d}x} \\ f(x) &= \int _{0}^{x}\frac {\left (\left (\left (\operatorname {csgn}\left (a \right ) a \,\tau ^{2}+\tau ^{2}+1\right ) \sqrt {a^{2} \left (\tau ^{2}+1\right )}+a \tau \left (\tau ^{2}+1\right ) \left (a +\operatorname {csgn}\left (a \right )\right )\right ) \left (a \tau \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (\tau ^{2}+1\right )}\right )^{\frac {-a -\operatorname {csgn}\left (a \right )}{a}}-\left (a \tau \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (\tau ^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} \left (\tau \sqrt {a^{2} \left (\tau ^{2}+1\right )}+\tau ^{2}+1\right )\right ) y}{\left (\tau ^{2}+1\right )^{{3}/{2}} \sqrt {a^{2} \left (\tau ^{2}+1\right )}}d \tau + c_1 \\ \end{align*}

Where \(c_1\) is constant of integration. Substituting result found above for \(f(x)\) into equation (3) gives \(\phi \)

\[ \phi = \frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} y}{\sqrt {x^{2}+1}}+ c_1 \]

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

\[ c_1 = \frac {\left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{-\frac {\operatorname {csgn}\left (a \right )}{a}} y}{\sqrt {x^{2}+1}} \]

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

\begin{align*} y = c_1 \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {\operatorname {csgn}\left (a \right )}{a}} \sqrt {x^{2}+1} \end{align*}

1.80.5 Solved using Lie symmetry for ﬁrst order ode

\begin{align*} y^{\prime }&=\frac {y \left (a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}\right )}{a^{2} \left (x^{2}+1\right )}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}

\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

\begin{equation} \tag{5E} b_{2}+\frac {y \left (a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}\right ) \left (b_{3}-a_{2}\right )}{a^{2} \left (x^{2}+1\right )}-\frac {y^{2} \left (a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{2} a_{3}}{a^{4} \left (x^{2}+1\right )^{2}}-\left (\frac {y \left (a^{2}+\frac {a^{2} x}{\sqrt {a^{2} \left (x^{2}+1\right )}}\right )}{a^{2} \left (x^{2}+1\right )}-\frac {2 y \left (a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}\right ) x}{a^{2} \left (x^{2}+1\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}\right ) \left (x b_{2}+y b_{3}+b_{1}\right )}{a^{2} \left (x^{2}+1\right )} = 0 \end{equation}

\[ -\frac {a^{4} x^{5} b_{2}+a^{4} x^{3} y^{2} a_{3}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{3} b_{1}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{2} y a_{1}+a^{4} x^{4} b_{1}-a^{4} x^{3} y a_{1}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{2} b_{2}+2 \sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x y a_{2}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} y^{2} a_{3}+2 a^{4} x^{3} b_{2}+a^{4} x^{2} y a_{2}+a^{4} x \,y^{2} a_{3}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x b_{1}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} y a_{1}+2 a^{4} x^{2} b_{1}-a^{4} x y a_{1}+\left (a^{2} \left (x^{2}+1\right )\right )^{{3}/{2}} y^{2} a_{3}-b_{2} a^{4} \sqrt {a^{2} \left (x^{2}+1\right )}+a^{4} x b_{2}+a^{4} y a_{2}+a^{4} b_{1}}{a^{4} \left (x^{2}+1\right )^{2} \sqrt {a^{2} \left (x^{2}+1\right )}} = 0 \]

\begin{equation} \tag{6E} -a^{4} x^{5} b_{2}-a^{4} x^{3} y^{2} a_{3}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{3} b_{1}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{2} y a_{1}-a^{4} x^{4} b_{1}+a^{4} x^{3} y a_{1}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{2} b_{2}-2 \sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x y a_{2}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} y^{2} a_{3}-2 a^{4} x^{3} b_{2}-a^{4} x^{2} y a_{2}-a^{4} x \,y^{2} a_{3}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x b_{1}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} y a_{1}-2 a^{4} x^{2} b_{1}+a^{4} x y a_{1}-\left (a^{2} \left (x^{2}+1\right )\right )^{{3}/{2}} y^{2} a_{3}+b_{2} a^{4} \sqrt {a^{2} \left (x^{2}+1\right )}-a^{4} x b_{2}-a^{4} y a_{2}-a^{4} b_{1} = 0 \end{equation}

\begin{equation} \tag{6E} -a^{4} x^{4} y a_{2}-a^{4} x^{3} y^{2} a_{3}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{3} b_{1}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{2} y a_{1}-a^{4} x^{3} y a_{1}-a^{4} \left (x^{2}+1\right ) x^{3} b_{2}+a^{4} \left (x^{2}+1\right ) x^{2} y a_{2}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x^{2} b_{2}-2 \sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x y a_{2}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} y^{2} a_{3}-a^{4} x^{2} y a_{2}-a^{4} x \,y^{2} a_{3}-a^{4} \left (x^{2}+1\right ) x^{2} b_{1}+2 a^{4} \left (x^{2}+1\right ) x y a_{1}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} x b_{1}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{4} y a_{1}-a^{4} x y a_{1}-\left (a^{2} \left (x^{2}+1\right )\right )^{{3}/{2}} y^{2} a_{3}-a^{4} \left (x^{2}+1\right ) x b_{2}-a^{4} \left (x^{2}+1\right ) y a_{2}+b_{2} a^{4} \sqrt {a^{2} \left (x^{2}+1\right )}-a^{4} \left (x^{2}+1\right ) b_{1} = 0 \end{equation}

\[ -a^{2} \left (a^{2} x^{5} b_{2}+a^{2} x^{3} y^{2} a_{3}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{2} x^{3} b_{1}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{2} x^{2} y a_{1}+a^{2} x^{4} b_{1}-a^{2} x^{3} y a_{1}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{2} x^{2} b_{2}+2 \sqrt {a^{2} \left (x^{2}+1\right )}\, a^{2} x y a_{2}+a^{2} \sqrt {a^{2} \left (x^{2}+1\right )}\, y^{2} a_{3}+\sqrt {a^{2} \left (x^{2}+1\right )}\, x^{2} y^{2} a_{3}+2 a^{2} x^{3} b_{2}+a^{2} x^{2} y a_{2}+a^{2} x \,y^{2} a_{3}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{2} x b_{1}+\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{2} y a_{1}+2 a^{2} x^{2} b_{1}-a^{2} x y a_{1}-\sqrt {a^{2} \left (x^{2}+1\right )}\, a^{2} b_{2}+y^{2} a_{3} \sqrt {a^{2} \left (x^{2}+1\right )}+a^{2} x b_{2}+a^{2} y a_{2}+a^{2} b_{1}\right ) = 0 \]

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \left \{x, y, \sqrt {a^{2} \left (x^{2}+1\right )}\right \} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \left \{x = v_{1}, y = v_{2}, \sqrt {a^{2} \left (x^{2}+1\right )} = v_{3}\right \} \]

\begin{equation} \tag{7E} -a^{2} \left (a^{2} v_{1}^{3} v_{2}^{2} a_{3}+a^{2} v_{1}^{5} b_{2}-a^{2} v_{1}^{3} v_{2} a_{1}-v_{3} a^{2} v_{1}^{2} v_{2} a_{1}+a^{2} v_{1}^{4} b_{1}+v_{3} a^{2} v_{1}^{3} b_{1}+a^{2} v_{1}^{2} v_{2} a_{2}+2 v_{3} a^{2} v_{1} v_{2} a_{2}+a^{2} v_{1} v_{2}^{2} a_{3}+a^{2} v_{3} v_{2}^{2} a_{3}+2 a^{2} v_{1}^{3} b_{2}-v_{3} a^{2} v_{1}^{2} b_{2}+v_{3} v_{1}^{2} v_{2}^{2} a_{3}-a^{2} v_{1} v_{2} a_{1}+v_{3} a^{2} v_{2} a_{1}+2 a^{2} v_{1}^{2} b_{1}+v_{3} a^{2} v_{1} b_{1}+a^{2} v_{2} a_{2}+a^{2} v_{1} b_{2}-v_{3} a^{2} b_{2}+v_{2}^{2} a_{3} v_{3}+a^{2} b_{1}\right ) = 0 \end{equation}

\[ \{v_{1}, v_{2}, v_{3}\} \]

\begin{equation} \tag{8E} -a^{4} b_{2} v_{1}^{5}-a^{4} b_{1} v_{1}^{4}-a^{4} a_{3} v_{1}^{3} v_{2}^{2}+a^{4} a_{1} v_{1}^{3} v_{2}-a^{4} b_{1} v_{1}^{3} v_{3}-2 a^{4} b_{2} v_{1}^{3}+a^{4} a_{1} v_{1}^{2} v_{2} v_{3}-a^{4} a_{2} v_{1}^{2} v_{2}+a^{4} b_{2} v_{1}^{2} v_{3}-a^{2} a_{3} v_{3} v_{1}^{2} v_{2}^{2}-2 a^{4} b_{1} v_{1}^{2}-2 a^{4} a_{2} v_{1} v_{2} v_{3}-a^{4} a_{3} v_{1} v_{2}^{2}+a^{4} a_{1} v_{1} v_{2}-a^{4} b_{1} v_{1} v_{3}-a^{4} b_{2} v_{1}-a^{2} \left (a^{2} a_{3}+a_{3}\right ) v_{2}^{2} v_{3}-a^{4} a_{1} v_{2} v_{3}-a^{4} a_{2} v_{2}+a^{4} b_{2} v_{3}-a^{4} b_{1} = 0 \end{equation}

Setting each coeﬃcients in (8E) to zero gives the following equations to solve

\begin{align*} a^{4} a_{1}&=0\\ a^{4} b_{2}&=0\\ -a^{2} a_{3}&=0\\ -a^{2} \left (a^{2} a_{3}+a_{3}\right )&=0\\ -a^{4} a_{1}&=0\\ -2 a^{4} a_{2}&=0\\ -a^{4} a_{2}&=0\\ -a^{4} a_{3}&=0\\ -2 a^{4} b_{1}&=0\\ -a^{4} b_{1}&=0\\ -2 a^{4} b_{2}&=0\\ -a^{4} b_{2}&=0 \end{align*}

\begin{align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= 0 \\ \eta &= y \\ \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the ﬁrst pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{y}} dy \end{align*}

\begin{align*} S&= \ln \left (y \right ) \end{align*}

Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,y) &= \frac {y \left (a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}\right )}{a^{2} \left (x^{2}+1\right )} \end{align*}

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= 0\\ S_{y} &= \frac {1}{y} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= \frac {a^{2} x +\sqrt {a^{2} \left (x^{2}+1\right )}}{a^{2} \left (x^{2}+1\right )}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= \frac {a^{2} R +\sqrt {a^{2} \left (R^{2}+1\right )}}{a^{2} \left (R^{2}+1\right )} \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {\frac {a^{2} R +\sqrt {a^{2} \left (R^{2}+1\right )}}{a^{2} \left (R^{2}+1\right )}\, dR}\\ S \left (R \right ) &= \frac {\ln \left (\frac {a^{2} R}{\sqrt {a^{2}}}+\sqrt {R^{2} a^{2}+a^{2}}\right )}{\sqrt {a^{2}}}+\frac {\ln \left (R^{2}+1\right )}{2} + c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \ln \left (y\right ) = \frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+a^{2}}\right )}{\sqrt {a^{2}}}+\frac {\ln \left (x^{2}+1\right )}{2}+c_2 \end{align*}

\begin{align*} y = {\mathrm e}^{\frac {\ln \left (x^{2}+1\right ) \sqrt {a^{2}}+2 c_2 \sqrt {a^{2}}+2 \ln \left (a^{2} x +\sqrt {a^{2} x^{2}+a^{2}}\, \sqrt {a^{2}}\right )-\ln \left (a^{2}\right )}{2 \sqrt {a^{2}}}} \end{align*}

1.80.6 Maple step by step solution

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

1.80.7 Maple trace

1.80.8 Maple dsolve solution

Solving time : 0.001 (sec)
Leaf size : 36

dsolve(diff(y(x),x) = y(x)*(1+a^2*x/(a^2*(x^2+1))^(1/2))/(a^2*(x^2+1))^(1/2), 
       y(x),singsol=all)

\[ y = c_1 \left (a x \,\operatorname {csgn}\left (a \right )+\sqrt {a^{2} \left (x^{2}+1\right )}\right )^{\frac {1}{\sqrt {a^{2}}}} \sqrt {x^{2}+1} \]

1.80.9 Mathematica DSolve solution

Solving time : 0.301 (sec)
Leaf size : 116

DSolve[{D[y[x],x]== y[x]*(1+ a^2*x/Sqrt[a^2*(x^2+1)])/Sqrt[a^2*(x^2+1)],{}}, 
       y[x],x,IncludeSingularSolutions->True]

\begin{align*} y(x)\to c_1 \left (a \left (-\sqrt {a^2 \left (x^2+1\right )}+\sqrt {a^2}+a x\right )\right )^{-\frac {a+1}{a}} \left (a \left (\sqrt {a^2 \left (x^2+1\right )}-\sqrt {a^2}+a x\right )\right )^{\frac {1}{a}-1} \left (\sqrt {a^2} \sqrt {a^2 \left (x^2+1\right )}-a^2 \left (x^2+1\right )\right ) \\ y(x)\to 0 \\ \end{align*}

1.80 problem 79

1.80.1 Solved as ﬁrst order linear ode