Question

$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{x}\mathit{y}\mathbf{=}\mathit{y}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathit{y}\mathbf{=}\mathbf{1}$when $\mathit{x}\mathbf{=}\mathbf{1}$.

Short answer

Answer:

The solution containing one arbitrary constant is $y=Cx{e}^x$, and with boundary condition, the value of constant is $C=\frac{1}{e}$. The particular solution is $x{\left(1+y^2\right)}^2=20$, and the plot of a slope field and some solution curves are as follows.

Step-by-step solution

Given information

The given differential equation is $xy\text{'}-xy=y$with boundaries $y=1$when$x=1$.

Definition of differential equation

A differential equation is a mathematical equation that connects one or more unknown functions with their derivatives.

Separate the variables

Separate the variables in $xy\text{'}-xy=y$and keep y on one side and x on the other.

$$\begin{gathered} x\frac{dy}{dx}-xy=y \\ xdy=\left(1+x\right)dx \\ \frac{1}{y}dy=\frac{1+x}{x}dx \\ \frac{1}{y}dy=\frac{1+x}{x}dx \end{gathered}$$

Integrate the differential equation

Integrate the separated form of differential equation with an arbitrary constant C.

$$\begin{gathered} \frac{1}{y}dy=\frac{1+x}{x}dx \\ \int \frac{1}{y}dy=\int \left(\frac{1}{x}+1\right)dx \\ Iny=Inx+x+c \\ Iny=Inx+In{e}^x+InC \end{gathered}$$

Solve further

$y=Cx{e}^x$ …… (1)

Find the value of the arbitrary constant and particular solution

Put the boundary conditions, $y=1$when $x=1$, in (1) and find the value of C.

$$\begin{gathered} \left(1\right)=C\left(1\right)e^{\left(1\right)} \\ 1=Ce \\ C=\frac{1}{e} \end{gathered}$$

Substitute the value of C in (1) and find the particular solution.

$y=x{e}^{x-1}$

Find the value of the arbitrary constant and particular solution

Put the boundary conditions, $y=1$when $x=1$, in (1) and find the value of C.

$$\begin{gathered} \left(1\right)=C\left(1\right)e^{\left(1\right)} \\ 1=Ce \\ C=\frac{1}{e} \end{gathered}$$

Substitute the value of C in (1) and find the particular solution.

$y=x{e}^{x-1}$

Plot the slope field

Plot the slope field of $xy\text{'}-xy=y$and some solution curves.

Therefore, for the differential equation $xy\text{'}-xy=y$, the solution containing one arbitrary constant is $y=Cx{e}^x$, and with boundary condition, $y=1$when $x=1$. The value of constant is $C=\frac{1}{e}$, and the particular solution is $y=x{e}^{x-1}$. The plot of a slope field and some solution curves are as follows.