For each of the following differential equations, separate variables and find a solution containing one arbitrary constant. Then find the value of the constant to give a particular solution satisfying the given boundary condition. Computer plot a slope field and some of the solution curves.

$\mathbf{10}\mathbf{.}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{xy}\mathbf{=}\mathbf{x}$ y = 1when x = 0

We have given a differential equation ${\mathrm{y}}^{\text{'}}-\mathrm{xy}=\mathrm{x}$with the boundary condition y = 1 when x = 0 .

Any equation of the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}$is called separable that is any equation in which dxand terms involving xcan be put on one side and, and terms involving yon other. For example,

f (x) dx = g (y) dy.

Any solution of the differential equation containing linearly independent arbitrary constant is the general solution of the differential equation.

Let’s first start by separating the variables

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}(\mathrm{y}+1) \\ \frac{\mathrm{dy}}{\mathrm{y}+1}=\mathrm{xdx} \end{gathered}$$

For the general solution we integrate both sides

$$\begin{gathered} \int \frac{\mathrm{dy}}{\mathrm{y}+1}=\int \mathrm{xdx} \\ \ln (\mathrm{y}+1)=\frac{{\mathrm{x}}^2}{2}+\mathrm{C} \end{gathered}$$

So, the general solution is

localid="1659245695144" $\mathrm{y}={\mathrm{C}}_1{\mathrm{e}}^{\raisebox{1ex}{${\mathrm{x}}^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-1$

The solution obtained from the general solution by giving some particular values to the arbitrary constants is called particular solution.

The value of the constant $C_1$when the boundary condition y = 1 when x = 1 is satisfied is

$$\begin{gathered} 1={\mathrm{C}}_1-1 \\ {\mathrm{C}}_1=2 \end{gathered}$$

The particular solution is

role="math" localid="1659245430520" $\mathrm{y}=2{\mathrm{e}}^{\raisebox{1ex}{${\mathrm{x}}^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-1$

Draw the slope field for this we will use the slope of the equation, which is

$\mathrm{y}\text{'}=\mathrm{x}(\mathrm{y}+1)$