3.3.2.1 Examples

3.3.2.1.1 Example 1 Solve

\begin{equation} y^{\prime }=B+Cf\left ( ax+by+c\right ) \tag {1}\end{equation}

This form of ode can be solved by letting \(u=ax+by+c\) which makes the ode separable.

\[ \frac {du}{dx}=a+by^{\prime }\]

Or

\[ y^{\prime }=\frac {u^{\prime }-a}{b}\]

The ode becomes

\begin{align*} \frac {u^{\prime }-a}{b} & =B+CF\left ( u\right ) \\ u^{\prime } & =bB+bCF\left ( u\right ) +a\\ \frac {du}{bB+bCF\left ( u\right ) +a} & =dx \end{align*}

Integrating gives

\begin{align*} \int \frac {du}{bB+bCF\left ( u\right ) +a} & =x+c\\ \int ^{ax+by+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a} & =x+c \end{align*}

If initial conditions are given as \(y\left ( x_{0}\right ) =y_{0}\), the above becomes

\begin{align*} \int _{0}^{ax_{0}+by_{0}+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a} & =x_{0}+c_{1}\\ c_{1} & =\int _{0}^{ax_{0}+by_{0}+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a}-x_{0}\end{align*}

Substituting this into (2) gives

\[ \int _{0}^{ax+by+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a}=x+\int _{0}^{ax_{0}+by_{0}+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a}-x_{0}\]

Note that when IC are given, the integrals are changed to have lower limit start from zero. If no initial conditions are given, lower limit is not used. This uses Maple’s Intat notation for integral at a point notation. See Maple help for Intat command.

3.3.2.1.2 Example 2 Solve

\begin{align*} y^{\prime } & =\frac {1}{7}F\left ( 3x+5y\right ) \\ y\left ( x_{0}\right ) & =y_{0}\end{align*}

Comparing the above to (1) shows that

\begin{align*} B & =0\\ C & =\frac {1}{7}\\ a & =3\\ b & =5\\ c & =0 \end{align*}

Plugging these into (2) gives

\begin{align*} \int ^{ax+by+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a} & =x+c_{1}\\ \int ^{3x+5y}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3} & =x+c_{1}\end{align*}

Applying IC gives

\[ \int _{0}^{3x_{0}+5y_{0}}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3}=c_{1}\]

Hence the solution is

\[ \int _{0}^{3x+5y}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3}=x+\int _{0}^{3x_{0}+5y_{0}}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3}\]

If IC were given as \(y\left ( 0\right ) =0\) then we see that \(c_{1}=0\) because upper limit becomes zero and the above solution becomes

\[ \int _{0}^{3x+5y}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3}=x \]

3.3.2.1.3 Example 3

\[ y^{\prime }=\sin \left ( 3x+5y\right ) \]

Comparing the above to (1) shows that

\begin{align*} B & =0\\ C & =1\\ a & =3\\ b & =5\\ c & =0 \end{align*}

Plugging these into (2) gives

\begin{align*} \int ^{ax+by+c}\frac {d\tau }{bB+bC\sin \left ( \tau \right ) +a} & =x+c_{1}\\ \int ^{3x+5y}\frac {d\tau }{5\sin \left ( \tau \right ) +3} & =x+c_{1}\end{align*}

3.3.2.1.4 Example 4

\[ y^{\prime }=8+3F\left ( 3x+5y+9\right ) \]

Comparing the above to (1) shows that

\begin{align*} B & =8\\ C & =3\\ a & =3\\ b & =5\\ c & =9 \end{align*}

Plugging these into (2) gives

\begin{align*} \int ^{ax+by+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a} & =x+c_{1}\\ \int ^{3x+5y+9}\frac {d\tau }{40+15F\left ( \tau \right ) +3} & =x+c_{1}\end{align*}

3.3.2.1.5 Example 5 This method works only when the argument of \(F\left ( \cdot \right ) \) is linear in \(x\) and \(y\). Lets see why. Assuming the ode is

\[ y^{\prime }=F\left ( x^{2}+5y\right ) \]

Let \(u=x^{2}+5y\) then \(\frac {du}{dx}=2x+5y^{\prime }\). Hence \(y^{\prime }=\frac {u^{\prime }-2x}{5}\) and the ode becomes

\begin{align*} \frac {u^{\prime }-2x}{5} & =F\left ( u\right ) \\ u^{\prime } & =5F\left ( u\right ) +2x \end{align*}

Which is no longer separable. Lets see what happens if \(y\) was not linear. Let the ode be

\[ y^{\prime }=F\left ( x+y^{2}\right ) \]

Let \(u=x+y^{2}\) then \(\frac {du}{dx}=1+2yy^{\prime }\). Hence \(y^{\prime }=\frac {u^{\prime }-1}{2y}\) and the ode becomes

\begin{align*} \frac {u^{\prime }-1}{2y} & =F\left ( u\right ) \\ u^{\prime } & =2yF\left ( u\right ) +1 \end{align*}

We see that the term \(y\) did not vanish and this can not work. This shows that for this method to work, the argument of the function \(F\) must be linear in \(x,y\)