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.

$\mathbf{11}\mathbf{.}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}{\mathbf{(}\mathit{y}\mathbf{-}\mathbf{2}\mathbf{)}}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.}$ y = 3when x = 1

We have given a differential equation $2\mathrm{y}\text{'}=3{(\mathrm{y}-2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$with the boundary condition y = 3 when x = 1.

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 dx and terms involving xcan be put on one side and dy, 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

$$\begin{gathered} 2\frac{\mathrm{dy}}{\mathrm{dx}}=3{\left(\mathrm{y}-2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ 2\frac{\mathrm{dy}}{{\left(\mathrm{y}-2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=3\mathrm{dx} \end{gathered}$$

For the general solution we integrate both sides

$$\begin{gathered} 2\int \frac{\mathrm{dy}}{{\left(\mathrm{y}-2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\int 3\mathrm{dx} \\ 3{\left(\mathrm{y}-2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=3\mathrm{x}+\mathrm{C} \end{gathered}$$

So, the general solution is

$y={(x+C_1)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2$

Where$C_1=\frac{C}{3}$

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 ${\mathrm{C}}_1$when the boundary condition y = 1 when x = 1 is satisfied is

$3={(1+{\mathrm{C}}_1)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\mathrm{C}}_1=0$

The particular solution is

$y={\left(x\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2$

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

$y\text{'}=\frac{3}{2}{(y-2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$