Examples

Example 1 The first step is to see if we can write the above as

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1}\end{equation}

Hence

\begin{equation} y^{\prime }=\frac {y}{x}-\frac {2}{x}e^{\frac {-y}{x}} \tag {2}\end{equation}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =1\\ g\left ( x\right ) & =-\frac {2}{x}\\ b & =-1\\ f\left ( b\frac {y}{x}\right ) & =e^{-\frac {y}{x}}\end{align*}

Hence the solution is

\begin{equation} y=ux \tag {A}\end{equation}

Where \(u\) is the solution to

\begin{equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) \tag {3}\end{equation}

Therefore \(f\left ( bu\right ) =e^{-u}\) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=-\frac {2}{x^{2}}e^{-u}\]

This is separable.

\begin{align*} e^{u}du & =-\frac {2}{x^{2}}dx\\ \int e^{u}du & =-2\int \frac {1}{x^{2}}dx\\ e^{u} & =\frac {2}{x}+c_{1}\\ u & =\ln \left ( \frac {2}{x}+c_{1}\right ) \end{align*}

Hence (A) becomes

\[ y=x\ln \left ( \frac {2}{x}+c_{1}\right ) \]

Example 2 Solve

\[ y^{\prime }x-y-2e^{x-\frac {y}{x}}=0 \]

The first step is to see if the above can be written as

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1}\end{equation}

Or

\begin{align} y^{\prime }x-y-2e^{x\frac {-y}{x}} & =0\nonumber \\ y^{\prime } & =\frac {y}{x}-\frac {2}{x}e^{x}e^{\frac {-y}{x}} \tag {2}\end{align}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =1\\ g\left ( x\right ) & =-\frac {2}{x}e^{x}\\ b & =-1\\ f\left ( b\frac {y}{x}\right ) & =e^{-\frac {y}{x}}\end{align*}

Hence the solution is

\begin{equation} y=ux \tag {A}\end{equation}

Where \(u\) is the solution to

\begin{equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) \tag {3}\end{equation}

Therefore \(f\left ( u\right ) =e^{-u}\) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=-\frac {2}{x^{2}}e^{x}e^{-u}\]

This is separable.

\begin{align*} e^{u}du & =-\frac {2}{x^{2}}e^{x}dx\\ \int e^{u}du & =-2\int \frac {e^{x}}{x^{2}}dx\\ e^{u} & =-2\left ( -\frac {e^{x}}{x}+\operatorname {Ei}\left ( x\right ) \right ) +c_{1}\end{align*}

Where \(\operatorname {Ei}\left ( x\right ) \) is the exponential integral \(\operatorname {Ei}\left ( x\right ) =\int _{-x}^{\infty }\frac {e^{-t}}{t}dt\).  Hence

\[ u=\ln \left ( c_{1}-2\left ( -\frac {e^{x}}{x}+\operatorname {Ei}\left ( x\right ) \right ) \right ) \]

And (A) becomes

\[ y=x\ln \left ( c_{1}-2\left ( -\frac {e^{x}}{x}+\operatorname {Ei}\left ( x\right ) \right ) \right ) \]

Example 3 Solve

\[ y^{\prime }x-y-2\sin \left ( 3\frac {y}{x}\right ) =0 \]

The first step is to see if we can write the above as

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1}\end{equation}

Hence

\begin{align} y^{\prime }x-y-2\sin \left ( 3\frac {y}{x}\right ) & =0\nonumber \\ y^{\prime } & =\frac {y}{x}-\frac {2}{x}\sin \left ( 3\frac {y}{x}\right ) \tag {2}\end{align}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =1\\ g\left ( x\right ) & =-\frac {2}{x}\\ b & =3\\ f\left ( b\frac {y}{x}\right ) & =\sin \left ( 3\frac {y}{x}\right ) \end{align*}

Hence the solution is

\begin{equation} y=ux \tag {A}\end{equation}

Where \(u\) is the solution to

\begin{equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) \tag {3}\end{equation}

Therefore \(f\left ( u\right ) =\sin \left ( 3u\right ) \) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=-\frac {2}{x^{2}}\sin \left ( 3u\right ) \]

This is separable.

\begin{align*} \frac {1}{\sin \left ( 3u\right ) }du & =-\frac {2}{x^{2}}dx\\ \int \frac {1}{\sin \left ( 3u\right ) }du & =-2\int \frac {1}{x^{2}}dx\\ \frac {1}{3}\left ( \ln \sin \left ( \frac {3u}{2}\right ) -\ln \cos \left ( \frac {3u}{2}\right ) \right ) & =\frac {2}{x}+c_{1}\\ \ln \sin \left ( \frac {3u}{2}\right ) -\ln \cos \left ( \frac {3u}{2}\right ) & =-\frac {6}{x}+c_{2}\\ \ln \frac {\sin \left ( \frac {3u}{2}\right ) }{\cos \left ( \frac {3u}{2}\right ) } & =-\frac {6}{x}+c_{2}\\ \ln \tan \left ( \frac {3u}{2}\right ) & =-\frac {6}{x}+c_{2}\\ \tan \left ( \frac {3u}{2}\right ) & =c_{3}e^{-\frac {6}{x}}\\ \frac {3u}{2} & =\arctan \left ( c_{3}e^{-\frac {6}{x}}\right ) \\ u & =\frac {2}{3}\arctan \left ( c_{3}e^{-\frac {6}{x}}\right ) \end{align*}

And (A) becomes

\[ y=\frac {2}{3}x\arctan \left ( c_{3}e^{-\frac {6}{x}}\right ) \]

Example 4 Solve

\[ y^{\prime }=\frac {y}{x}-\frac {2}{x}\sqrt {\sin \left ( 3\frac {y}{x}\right ) }\]

The first step is to see if we can write the above as

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1}\end{equation}

Hence

\begin{equation} y^{\prime }=\frac {y}{x}-\frac {2}{x}\left ( \sin \left ( 3\frac {y}{x}\right ) \right ) ^{\frac {1}{2}} \tag {2}\end{equation}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =2\\ g\left ( x\right ) & =-\frac {2}{x}\\ b & =3\\ f\left ( b\frac {y}{x}\right ) & =\sin \left ( 3\frac {y}{x}\right ) \end{align*}

Hence the solution is

\begin{equation} y=ux \tag {A}\end{equation}

Where \(u\) is the solution to

\begin{equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) ^{\frac {1}{2}} \tag {3}\end{equation}

Therefore \(f\left ( u\right ) =\sin \left ( 3u\right ) \) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=-\frac {2}{x^{2}}\sin \left ( 3u\right ) ^{\frac {1}{2}}\]

This is separable.

\begin{align*} \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =-\frac {2}{x^{2}}dx\\ \int \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =-2\int \frac {1}{x^{2}}dx\\ \int \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =\frac {2}{x}+c_{1}\end{align*}

Leaving the integral as is, since it is too complicated to solve, then using \(y=ux\) where \(u\) is the solution of the above.

Example 5 Solve

\[ y-2x^{3}\tan \left ( \frac {y}{x}\right ) -y^{\prime }x=0 \]

The first step is to see if we can write the above as

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1}\end{equation}

Hence

\begin{align} y-2x^{3}\tan \left ( \frac {y}{x}\right ) -y^{\prime }x & =0\nonumber \\ y^{\prime }x & =y-2x^{3}\tan \left ( \frac {y}{x}\right ) \nonumber \\ y^{\prime } & =\frac {y}{x}-2x^{2}\tan \left ( \frac {y}{x}\right ) \tag {2}\end{align}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =1\\ g\left ( x\right ) & =-2x^{2}\\ b & =1\\ f\left ( b\frac {y}{x}\right ) & =\tan \left ( \frac {y}{x}\right ) \end{align*}

Hence the solution is

\begin{equation} y=ux \tag {A}\end{equation}

Where \(u\) is the solution to

\begin{equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) \tag {3}\end{equation}

Therefore \(f\left ( u\right ) =\tan u\) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=-2x\tan u \]

This is separable.

\begin{align*} \frac {1}{\tan }du & =-2xdx\\ \int \frac {1}{\tan }du & =-2\int xdx\\ \ln \left ( \sin u\right ) & =-x^{2}+c_{1}\\ \sin u & =c_{2}e^{-x^{2}}\\ u & =\arcsin \left ( c_{2}e^{-x^{2}}\right ) \end{align*}

Hence (A) becomes

\[ y=x\arcsin \left ( c_{2}e^{-x^{2}}\right ) \]

Example 6 Solve

\[ y^{\prime }=\frac {y}{x}+x\sin \left ( \frac {y}{x}\right ) \]

The first step is to see if we can write the above as

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1}\end{equation}

Hence

\begin{equation} y^{\prime }=\frac {y}{x}+x\sin \left ( \frac {y}{x}\right ) \tag {2}\end{equation}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =1\\ g\left ( x\right ) & =x\\ b & =1\\ f\left ( b\frac {y}{x}\right ) & =\sin \left ( \frac {y}{x}\right ) \end{align*}

Hence the solution is

\begin{equation} y=ux \tag {A}\end{equation}

Where \(u\) is the solution to

\begin{equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) \tag {3}\end{equation}

Therefore \(f\left ( u\right ) =\sin u\) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=\frac {1}{x}\left ( x\right ) \sin \left ( u\right ) \]

This is separable.

\begin{align*} \frac {1}{\sin u}du & =dx\\ \int \frac {1}{\sin u}du & =\int dx\\ \ln \sin \frac {u}{2}-\ln \cos \frac {u}{2} & =x+c_{1}\\ \ln \tan \frac {u}{2} & =x+c_{1}\\ \tan \frac {u}{2} & =c_{2}e^{x}\\ \frac {u}{2} & =\arctan \left ( c_{2}e^{x}\right ) \\ u & =2\arctan \left ( c_{2}e^{x}\right ) \end{align*}

Hence (A) becomes

\[ y=2x\arctan \left ( c_{2}e^{x}\right ) \]