MHB Obtaining Differential Equations with Solution

[cheatmenot]

(a) all circles tangent to x-axis?? with answer on the book [1+(y^{2})^2]^3=[yy''+1+(y')^2]^2

(b) all circles with fixed radius r and tangent to x-axis??

Thanks a bunch!

 

[MarkFL]

I have moved your thread here to our Differential Equations forum.

(a) Can you begin by giving the family of circles tangent to the $x$-axis?

 

[mathmari]

(a) all circles tangent to x-axis?? with answer on the book [1+(y^{2})^2]^3=[yy''+1+(y')^2]^2

(b) all circles with fixed radius r and tangent to x-axis??

Thanks a bunch!

(a) The equation of a circle with center at $(x_0,y_0)$ and radius $r$ is: $$(x-x_0)^2+(y-y_0)^2=r^2$$
Since the circles should tangent to the $x-$axis, $y_0= \pm r$.
Therefore,
$$(x-x_0)^2+(y \mp r)^2=r^2 $$
$x_0$ and $r$ are arbitrary constants and must not appear in ODE which generates the family of the circles. Since there are two arbitrary constants to eliminate, the ODE will be second order.

Differentiating both sides with respect to $x$ we have the following:
$$2(x-x_0)+2(y \mp r)y'=0 $$

Differentiating again with respect to $x$ we have the following:
$$2+2(y')^2+2(y \mp r)y''=0 $$

Now you have to use these equations to get rid of $x_0$ and $r$, or to get rid of $(x-x_0)$ and $(y \mp r)$ by expressing them in terms of $y$ and its derivatives.
(b)The equation of a circle with center at $(x_0,y_0)$ and fixed radius $r$ is: $$(x-x_0)^2+(y-y_0)^2=r^2$$
Since the circles should tangent to the $x-$axis, $y_0= \pm r$.
Therefore,
$$(x-x_0)^2+(y \mp r)^2=r^2 $$
$x_0$ is an arbitrary constant and must not appear in ODE which generates the family of the circles. Since there is only one arbitrary constant to eliminate, the ODE will be 1st order.

Differentiating both sides with respect to $x$ we have the following:
$$2(x-x_0)+2(y \mp r)y'=0 $$

Now you have to use these equations to get rid of $x_0$, or to get rid of $(x-x_0)$ by expressing this in terms of $r$ and of $y$ and its derivatives.

 

[cheatmenot]

(a) The equation of a circle with center at $(x_0,y_0)$ and radius $r$ is: $$(x-x_0)^2+(y-y_0)^2=r^2$$
Since the circles should tangent to the $x-$axis, $y_0= \pm r$.
Therefore,
$$(x-x_0)^2+(y \mp r)^2=r^2 $$
$x_0$ and $r$ are arbitrary constants and must not appear in ODE which generates the family of the circles. Since there are two arbitrary constants to eliminate, the ODE will be second order.

Differentiating both sides with respect to $x$ we have the following:
$$2(x-x_0)+2(y \mp r)y'=0 $$

Differentiating again with respect to $x$ we have the following:
$$2+2(y')^2+2(y \mp r)y''=0 $$

Now you have to use these equations to get rid of $x_0$ and $r$, or to get rid of $(x-x_0)$ and $(y \mp r)$ by expressing them in terms of $y$ and its derivatives.
(b)The equation of a circle with center at $(x_0,y_0)$ and fixed radius $r$ is: $$(x-x_0)^2+(y-y_0)^2=r^2$$
Since the circles should tangent to the $x-$axis, $y_0= \pm r$.
Therefore,
$$(x-x_0)^2+(y \mp r)^2=r^2 $$
$x_0$ is an arbitrary constant and must not appear in ODE which generates the family of the circles. Since there is only one arbitrary constant to eliminate, the ODE will be 1st order.

Differentiating both sides with respect to $x$ we have the following:
$$2(x-x_0)+2(y \mp r)y'=0 $$

Now you have to use these equations to get rid of $x_0$, or to get rid of $(x-x_0)$ by expressing this in terms of $r$ and of $y$ and its derivatives.

thank you for your answer but as i go over and answer them, i got it really hard in obtaining it.. can you do it for me sir?? i really needed it thanks a lot . .more power i need both ..

 

[mathmari]

(a) We have the following equations:

$$(x-x_0)^2+(y \mp r)^2=r^2 \ \ \ (1)$$
$$2(x-x_0)+2(y \mp r)y'=0 \ \ \ (2)$$
$$2+2(y')^2+2(y \mp r)y''=0 \ \ \ (3)$$

Since we want to get get rid of $(x-x_0)$ and $(y \mp r)$ we do the following:

At the equation $(3)$ solve for $(y \mp r)$ and replace it at the equations $(1)$ and $(2)$.
Then solve at the equation $(2)$ for $(x-x_0)$ and relace this at the equation $(1)$. What do you get??(b)We have the following equations:

$$(x-x_0)^2+(y \mp r)^2=r^2 \ \ \ (*)$$
$$2(x-x_0)+2(y \mp r)y'=0 \ \ \ (**)$$

Since we want to get rid of $(x-x_0)$ we do the following:

At the equation $(**)$ solve for $(x-x_0)$ and replace it at the equation $(*)$. What do you get??