widdonod1t

2021-12-27

Differential Equations (Families of Curves): Circles with fixed radius r and tangent to the y-axis.

poleglit3

Beginner2021-12-28Added 32 answers

The general equation of a circle with centre at (h, k) and radius r is given by

$(x-h)}^{2}+{(y-k)}^{2}={r}^{2$

Also, it is given that y axis is tangent to this circle.

This means the perpendicular distance from the centre of the circle to the line$x=0$ is equal to the radius of the circle.

This means$h=r$

Hence, the equation of such a circle becomes

$(x-r)}^{2}+{(y-k)}^{2}={r}^{2$

Here, k is hte only variable left.

Hence, to obtain the differential equation, we need to eliminate the variable k

We will do this with the help of differentiation.

Differentiating the curve with respect to x, we get

$2(x-r)+2(y-k)\frac{dy}{dx}=0$

$(y-k)\frac{dy}{dx}=r-x$

$(y-k)=\frac{r-x}{\frac{dy}{dx}}$

Putting the value of$y-k$ back in the equation we get

$(x-r)}^{2}+{(y-k)}^{2}={r}^{2$

$(x-r)}^{2}+\left(\frac{r-x}{\frac{dy}{dx}}\right)={r}^{2$

${\left(\frac{dy}{dx}\right)}^{2}{(x-r)}^{2}+{(x-r)}^{2}={\left(\frac{dy}{dx}\right)}^{2}\left({r}^{2}\right)$

${\left(\frac{dy}{dx}\right)}^{2}({x}^{2}+{r}^{2}-2xr-{r}^{2})+{(x-r)}^{2}=0$

${\left(\frac{dy}{dx}\right)}^{2}({x}^{2}-2xr)+{(x-r)}^{2}=0$

$\left(\frac{dy}{dx}\right)}^{2}=\frac{{(x-r)}^{2}}{2xr-{x}^{2}$

Also, it is given that y axis is tangent to this circle.

This means the perpendicular distance from the centre of the circle to the line

This means

Hence, the equation of such a circle becomes

Here, k is hte only variable left.

Hence, to obtain the differential equation, we need to eliminate the variable k

We will do this with the help of differentiation.

Differentiating the curve with respect to x, we get

Putting the value of

Marcus Herman

Beginner2021-12-29Added 41 answers

Vasquez

Expert2022-01-09Added 669 answers

Center is (0,k) and radius is r.

Equation is

By differentiating we get,

Putting the value of K from Equation 1 and 2, we get

star233

Skilled2023-05-13Added 403 answers

alenahelenash

Expert2023-05-13Added 556 answers

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$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$