Question

Find the family of curves satisfying the differential equation $\left(x+y\right)\mathit{d}\mathit{y}\mathbf{+}\left(x-y\right)\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}$and also find their orthogonal trajectories.

Short answer

Answer

$\left(0,0\right)$is the center of the circle and $\sqrt{2c_1}$is the radius of the circle. The value of $c_1$varies from $\infty$to $\infty$.

Step-by-step solution

Given information

The given differential equation is $\left(x+y\right)dy+\left(x-y\right)dx=0$

Standard equation of a circle

The standard equation of a circle is ${\left(x-h\right)}^{\mathbf{2}}\mathbf{+}{\left(y-k\right)}^{\mathbf{2}}\mathbf{=}{\mathit{r}}^{\mathbf{2}}$

Here, $\left(h,k\right)$is the center of the circle and r is the radius of the circle.

Slope of the equation

The given differential equation is,

$\left(x+y\right)dy+\left(x-y\right)dx=0$

Now, simplify the equation as,

$$\begin{gathered} \left(x+y\right)\mathit{d}\mathit{y}\mathbf{+}\left(x-y\right)\mathit{d}\mathit{x}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(x+y\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{-}\left(x-y\right)\mathit{d}\mathit{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{y}\mathbf{-}\mathbf{x}}{\mathbf{x}\mathbf{+}\mathbf{y}} \end{gathered}$$

This is the slope of the equation.

Slope of the orthogonal trajectory

The slope of the orthogonal trajectory is equal to the negative reciprocal of the slope of the equation.

Therefore, the slope of the orthogonal trajectory is,

$$\begin{gathered} \frac{dy}{dx}=-\frac{x+y}{y-x} \\ \frac{dy}{dx}=\frac{x+y}{x-y} \end{gathered}$$

Again, simplify the equation as,

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

Step 5: Integrate the equation

Now, integrate both sides of the equation as,

$$\begin{gathered} \int \left(x-y\right)dy=\int \left(x+y\right)dx \\ xy-\frac{y^2}{2}=\frac{x^2}{2}+xy+c \\ \frac{2^2}{2}+\frac{y^2}{2}=-c \\ \frac{x^2}{2}+\frac{y^2}{2}=c_1 \end{gathered}$$

$x^2+x={}^{2}2c{}_1$.....(1)

The standard equation of a circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2,$where $\left(h,k\right)$the center of the circle and r is the radius of the circle.

Now, compare equation (1) with the standard form of equation, which indicates that

$\left(0,0\right)$is the center of the circle and $\sqrt{2c_1}$is the radius of the circle. The value of $c_1$varies from $\infty$to $\infty$.