4.4.2.5 Example 5

Let solve the same ode above but with only one IC is given and not both. In other words, if we are given either $y\left(x_0\right)=y_0$ or $y^{\prime }\left(x_0\right)=y_0^{\prime }$ only. To see how to handle this method in such case. We know if there are no IC are given, then we use EQ (4) above, which is

$$\begin{array}{}\text{(4)}& {\int }_0^x{a}_0(\alpha ,y,y^{\prime })d\alpha +{\int }_0^y{a}_1(0,\beta ,y^{\prime })d\beta +{\int }_0^{y^{\prime }}{a}_2(0,0,\gamma )d\gamma =c_1\end{array}$$

And if both initial conditions are given, then we use EQ (3), which is

$$\begin{array}{}\text{(3)}& {\int }_{x_0}^x{a}_0(\alpha ,y,y^{\prime })d\alpha +{\int }_{y_0}^y{a}_1(x_0,\beta ,y^{\prime })d\beta +{\int }_{y_0^{\prime }}^{y^{\prime }}{a}_2(x_0,y_0,\gamma )d\gamma =0\end{array}$$

Let see what to do when only one IC is given for the second order ode

$$\begin{array}{rl}y{y}^{\prime \prime }+y^{\prime }(y^{\prime }-1)& =0\\ y\left(0\right)& =0\end{array}$$

From problem 4, we found that this match works

$$\begin{array}{rl}\text{(2A)}& a_2& =y{a}_1& =(y^{\prime }-1)\\ a_0& =0\end{array}$$

And now we are given $x_0,y_0$ only but we are not given $y_0^{\prime }$. Because of this, we will use (3) and not (4) and use the values for the given $x_0,y_0$ where needed and replace $y_0^{\prime }$ by $y^{\prime }\left(0\right)$. Hence (3) becomes

$$\begin{array}{rl}{\int }_{x_0}^x{a}_0(\alpha ,y,y^{\prime })d\alpha +{\int }_{y_0}^y{a}_1(x_0,\beta ,y^{\prime })d\beta +{\int }_{y^{\prime }\left(0\right)}^{y^{\prime }}{a}_2(x_0,y_0,\gamma )d\gamma & =0\\ 0+{\int }_0^y(y^{\prime }-1)d\beta +{\int }_{y^{\prime }\left(0\right)}^{y^{\prime }}{y}_{0}d\gamma & =0\\ 0+(y^{\prime }-1)y+{\int }_{y^{\prime }\left(0\right)}^{y^{\prime }}\left(0\right)d\gamma & =0\\ (y^{\prime }-1)y& =0\end{array}$$

Hence $y=0$ or $y^{\prime }=1$. But $y^{\prime }=1$. Solving this gives $y=x+c_1$. using initial conditions $y\left(0\right)=0$ gives $c=0$. Hence $y=x$ is also a solution. Hence solutions are

$$\begin{array}{rl}y& =0\\ y& =x\end{array}$$

This shows that if we are given even partial initial conditions, then we should use EQ (3) and not EQ (4). The following example gives one more illustration of this.